
Every point (x,y) on the curve y=log(3x)/log2 is transferred to a new point by the following translation (x′,y′)=(x−m,y−n), where m and n are integers. The set of (x′,y′) form the curve y=log(12x−96)/log2. What is

Every point (x,y) on the curve y=log23x is transferred to a new point by the following translation (x′,y′)=(x+m,y+n), where m and n are integers. The set of (x′,y′) form the curve y=log2(12x−96). What is the value of m+n?

I'm working on logarithmic equations and I'm stuck on how my book arrives at the next step. First, they use the change of base formula on, log(sqrt(2))(x^3  2) (sqrt(2)) is the base,changing to base 2 log(sqrt(2))(x^3  2)= log2(x^3  2)/(log2(sqrt(2)) I

1. The sequence log2 32, log2 y, log2 128, ... forms an arithmetic sequence. What is the value of y? 2. If log a^2 b^3 = x and log (a/b) = y, what are the values of log a and log b?

Every point (x,y) on the curve y = \log_{2}{3x} is transferred to a new point by the following translation (x',y') =(x+m,y+n), where m and n are integers. The set of (x',y') form the curve y = \log_{2}{(12x96)} . What is the value of m + n ?


I don't understand how to do these w/o calc. I tried to write it in a way that will make someone understand how to read it. Hope I typed it clear enough.Thanks so much for the help anyone! How to find the exact value of logarithm: 10. log5^100 log5^4 11.

The problem I have to solve is log with base 2 ^6 multiply by log base 6 ^ 8. I use the change of base formula and got log6/log2 * log8/log6 Which become log6/log2 * log2()^3/ log 6 I'm stuck here thanks.

A curve passes through the point (1,11) and it's gradient at any point is ax^2 + b, where a and b are constants. The tangent to the curve at the point (2,16) is parallel to the xaxis. Find i) the values of a and b ii) the equation of the curve

the tangent yo the curve y=x^2 +5x 2 @ the point (1,4)intersect the normal to the same curve @ the point (3,8) at the point P.Find the coordinates of point P.[ans: 1/3,16/3] just give me some hint to calculate this solution.

1.Use the laws of logarithms to express log2 (6) â€“ log2 (3) + 2log2 (8)^1/2 as a single logarithm; then evaluate. 2. For logy = log(0.5x â€“ 3) + log2, express y as a function of x.

Solve: 2^(5x6) = 7 My work: log^(5x6) = log7 5x  6(log2) = log7 5x = log7 + 6(log2) x = (log7 + log2^6) / 5 And textbook answer: (log7) / (log2) What did I do wrong?

which three statements are true? a) if x= 10^4 then log 10 = 4 b)if x= 2^8 then log 2x = 8 c) log2 2= 4 d) if x= 3 then log10 3=x e) log 10 2562log 10 a/log 10 b f)log 10 (ab)= log 10 a/log 10 b g) the gradient of the graph of y= 2x^x at x= 2 is 2e^e

Given that x²cos ysin y=0 ,(0,π): a)verfiy that given point is on the curve. b)use implicit differentiation to find the slope of the above curve at the given point. c)find the equation for tangent and normal to the curve at that point.

which three statements are true? a) if x= 10^4 then log 10 = 4 b)if x= 2^8 then log 2x = 8 c) log2 2= 4 d) if x= 3 then log10 3=x e) log 10 2562log 10 a/log 10 b f)log 10 (ab)= log 10 a/log 10 b g) the gradient of the graph of y= 2x^x at x= 2 is 2e^e

which three statements are true? a) if x= 10^4 then log 10 = 4 b)if x= 2^8 then log 2x = 8 c) log2 2= 4 d) if x= 3 then log10 3=x e) log 10 2562log 10 a/log 10 b f)log 10 (ab)= log 10 a/log 10 b g) the gradient of the graph of y= 2x^x at x= 2 is 2e^e


A racecar is initially travelling at 75 mph at point A as it enters the Scurve shown. In order to successfully traverse the curve, the racecar driver applies his brakes and decelerates uniformly between point A and B. Point B is located 750 ft down the

A racecar is initially travelling at 75 mph at point A as it enters the Scurve shown. In order to successfully traverse the curve, the racecar driver applies his brakes and decelerates uniformly between point A and B. Point B is located 750 ft down the

The slope of the tangent line to a curve at any point (x, y) on the curve is x/y. What is the equation of the curve if (4, 1) is a point on the curve? x2 − y2 = 15 x2 + y2 = 15 x + y = 15 < my answer xy = 15

which three statements are true? a) if x= 10^4 then log 10 = 4 b)if x= 2^8 then log 2x = 8 c) log2 2= 4 d) if x= 3 then log10 3=x e) log 10 2562log 10 a/log 10 b f)log 10 (ab)= log 10 a/log 10 b g) the gradient of the graph of y= 2x^x at x= 2 is 2e^e

1. a.) Find an equation for the line perpendicular to the tangent curve y=x^3  9x + 5 at the point (3,5) [* for a. the answer that I obtained was y5 = 1/18 (x3) ] b.) What is the smallest slope on the curve? At what point on the curve does the curve

Hello! Could someone please take a look at the problem below and let me know if I made mistakes in simplifying the given equation? I'm quite certain that I did make a mistake somewhere because the original equation and its simplified form give different

The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (3, 1) is a point on the curve?

The slope of the tangent line to a curve at any point (x, y) on the curve is x/y. What is the equation of the curve if (4, 1) is a point on the curve? x2 − y2 = 15 x2 + y2 = 15 x + y = 15 xy = 15

The slope of the tangent line to a curve at any point (x, y) on the curve is x divided by y. What is the equation of the curve if (2, 1) is a point on the curve?

If u = log(r), where r^2 = (xa)^2 + (yb)^2, and (x1) and (yb) are not zero simultaneously, show that d^2u/dx^2 + d^2u/dy^2 = 0. I first used some of the properties of log and made u = 1/2 * log((xa)^2 + (yb)^2) Then made u = 1/2 * ln((xa)^2 +


Tell what the output value is for the function machine for the given values. log 16.9=1.23 log2^152=9 log2^1=0 log0.046=1.34 thank you

how to solve for radius of horizontal curve with coordinates(NE) for points A,B, C on the curve Point A : N1405.4018 E1256.7569 Point B : N1283.3703 E1294.7027 Point C : N1225.9373 E1286.6137

how to solve for radius of horizontal curve with coordinates(NE) for points A,B, C on the curve Point A : N1405.4018 E1256.7569 Point B : N1283.3703 E1294.7027 Point C : N1225.9373 E1286.6137

The slope of the tangent to a curve at any point (x, y) on the curve is x/y . Find the equation of the curve if the point (3,4) on the curve.

More log/exponential equations log2 x + log2(x+2) =3

consider the curve defined by the equation y=a(x^2)+bx+c. Take a point(h,k) on the curve. use Wallis's method of tangents to show that the slope of the line tangent to this curve at the point(h,k) will be m= 2ah+b. have to prove this for tow cases: a>0

1. Log10²x+log10x²=log10² 21 2. Log4(log2x)+log2(log4x)=2 3. X^logx+5/3= 10^5+log x 4. Log 1/2(x1)+ log 1/2(x+1)log1/√2(7x)=1

I don't understand how log2 √(1/2) turned into log2 2^(1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A  logk B 3. logk (A^n) = n logk A where k is any positive number , k

I don't understand how log2 √(1/2) turned into log2 2^(1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A  logk B 3. logk (A^n) = n logk A where k is any positive number , k

I don't understand how log2 √(1/2) turned into log2 2^(1/2). Quote: You will have to know the 3 prime properties of logs 1. logk (AB) = logk A + logk B 2. logk(A/B) = logk A  logk B 3. logk (A^n) = n logk A where k is any positive number , k


How should i do this question?? it says: find the equation of the tangent to the curve y=(x2)^3 at the point (3,1). calculate the coordinates of the point where this tangent meets the curve again. I know how to get the first part, if i'm not wrong, it's y

If ln a=2, ln b=3, and ln c=5, evaluate the following: (a) ln(a^−2/b^4c^−3)= .9524 (b) ln(√b^−4*c^1*a^−1)= 1.739 (c) ln(a^2b^4)/ln(bc)^−2= 42.39 (d) (ln(c^−1))*(ln(a/b^3))^−4= .03507 I am getting these answers but the problem gives me an

Point A is 5 m from a loudspeaker. At point B, the loudspeaker sounds half as loud as at point A. How far is point B from the loudspeaker? I get 50 m, but the answer is supposed to be 15.8 m. I need to know what I'm doing wrong. By my (apparently flawed)

a curve ahs parametric equations x=t^2 and y= 11/2t for t>0. i)find the coordinates of the point P where the curve cuts the xaxis which i found to be P(1/4, 0) the next part i cant do ii) find the gradient of the curve at this point. So far, I have

given that log2 3 = x, log 2 5=y and log 2 7 = z, express log 2 21 in terms of x,y, and z the 2 is the base

A curve is traced by a point P(x,y) which moves such that its distance from the point A(1,1) is three times the distance from the point B(2,1). Find the equation of the curve and identity.

1. Given the curve a. Find an expression for the slope of the curve at any point (x, y) on the curve. b. Write an equation for the line tangent to the curve at the point (2, 1) c. Find the coordinates of all other points on this curve with slope equal to

i need help with these two homework problems Use the Laws of Logarithms to combine the expression into a single logarithm log2 5 − 5 log2 x + 1/2 log2(x + 1) Solve the logarithmic equation for x log2(x + 2) + log2(x − 1) = 2

Find the line which passes through the point (0, 1/4) and is tangent to the curve y=x^3 at some point. So I found the derivative which is 3x^2. Let (a, a3) be the point of tangency. 3x^2 = (a3  1/4)/(a0) I'm not sure how to solve for a. Yes, the point is

Consider the curve given by x^2+4y^2=7+3xy a) Show that dy/dx=(3y2x)/(8y3x) b) Show that there is a point P with xcoordinate 3 at which the line tangent to the curve at P is horizontal. Find the ycoordinate of P. c) Find the value of d^2y/dx^2 (second


A curve is traced by a point P(x,y) which moves such that its distance from the point A(1,1) is three times the distance from the point B(2,1). Find the equation of the curve and identity.

for the parametric curve defined by x=32t^2 and y=52t ...sketch the curve using the parametric equation to plot of the point. use an arrow to indicate the direction of the curve for o<t<1. Find an equation for the line tangent to the curve at the

Find an equation of the tangent to the curve at the given point. y=4(sinx)^2 point: pi/6,1) So I took the derivative of the original function to get: y' = 8cosx*sinx I then chose a point to plug in to find a point for the slope. i picked pi/6 because i

using logarithms to solve exponential equations. 5^1+x = 2^1x I need exact numbers. I did one on my own already. 5^x1 = 9 5^x1 = 9 log(5^x1) = log9 (log5)(x1) = log9 x1 = (log9/log5) x= (log9/log5)1 x = 2.3652 Logarithm help  Joe, Friday, September

When looking at a titration curve, I have to determine which is not true and I have it narrowed down to two options: The initial starting point on the titration curve is where pH depends only on [HA]0 or The finial point on a titration curve the pH depends

1. Solve the following simultaneous equations: log2 xy = 7 log2 (x^2/y) = 5 2. If log y x =a and log z x=b where x is not equal to 1, express the following in terms of a and b: logy (yz) Solve the following simultaneous equations: y = 2 log3 x y+1 = log3

A curve passes through the point (0, 2) and has the property that the slope of the curve at every point P is three times the ycoordinate of P. Find an equation of the curve. dy/dp = 3y or ∫ (3/y) dy = ∫ dp or 3 ln(y) = p + c or @ (0,2) ln(2) =

label a point f inside the curve. why is this an inefficient point? label a point g outside the curve. why is this point unattainable? why are pointS A THROUGH E ALL EFFICIENT POINTS?

label a point f inside the curve. why is this an inefficient point? label a point g outside the curve. why is this point unattainable? why are pointS A THROUGH E ALL EDDICIENT POINTS?

Solve the logarithmic equation for x. (Enter your answers as a commaseparated list.) log2(x + 17) − log2(x − 2) = 1 2 is the base for the log.


A curve passes through the point (7,6) and has the property that the slope of the curve at every point P is 4 times the ycoordinate of P. What is the equation of the curve? Simplify the equation as much as possible.

which 3 are correct a) if x= 10^4 then log10 x = 4 b)if x= 2^8 then log 2x = 8 c) log2 2= 4 d) if x= 3 then log10 3=x e) log 10 2562log 10 16=0 f)log 10 (ab)= log 10 a/log 10 b g) the gradient of the graph of y= 2e^x at x= 2 is 2e^2 h) the gradient of

A certain density curve looks like an interverted letter V. The first segment goes fro the point (0,0.6) to the point (0.5,1.4). The segment goes from (0.5.1.44) to (1,0.6). (a) Sketch the curve. Verify that the area under the curve is 1, so that it is a

Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2 + 1 and connect the tangent point to the xaxis. If the tangent point is close to the yaxis, the line segment is long. If the tangent point is far from the

The equation of a curve is y = 2x^3 + 3x^2 Find: xintercept of the curve yintercept of the curve b) Determine the stationery point of the curve. i) for each point in(b) above, determine whether it is a maximum or a minimum

find the eqt. of tangent to the curve y=x^2+2x10 @ the point where the curve cuts the yaxis.[ans:y=2x10] how do i do this because they didn't give me the point...?

find the coordinates of the point where the tangent to the curve y=x^3 +x +2 at the point (1,4) meets the curve again. [ans:2,8] pls help me i don't understand the question....

Consider the curve deﬁned by the equation y = 4x^3 +3x. Set up an integral that represents the length of curve from the point (0,0) to the point (4,268).

Find the xyequation of the curve that passes through (2, 2) and whose slope at any point on the curve is equal to 5 times the xcoordinate of that point

Logarithm!!! Select all of the following that are true statements: (a) log(2x) = log(2) + log(x) (b) log(3x) = 3 log(x) (c) log(12y) = 2 log(2) + log(3y) (d) log(5y) = log(20y) – log(4) (e) log(x) = log(5x) – log(5) (f) ln(25) = 2ln(5) (g) ln(1) =


solve the equation log2(x+4)log4x=2 the 2 and 4 are lower than the g This is what I got: log2(x+4)+log2(4^x)=2 log2((x+4)*4^x)=2 4^x(x+4)=4 x=0 is a solution???

Determine the equation of a curve in the xyplane that passes through the point (0, 1) and has the slope x2 sin 4x at any point (x, y) on the curve.

Find the coordinates of a point on the curve y=6x^2 at which the tangent of the curve at the point is perpendicular to the line y=1/4x+1.

The tangent to the curve 2y= 2x^2 5x +4 at the point where x=1 is parallel to the normal to the curve y= ax^2 + bx +10 at the point (2,2). Calculate the values of a and b. The answers are a=1, b=6

When looking at a titration curve, I have to determine which is false and I have it narrowed down to two options: The initial starting point on the titration curve is where pH depends only on [HA]0 or The finial point on a titration curve the pH depends

Show that the tangent line to the curve y=x^3 at the point x=a also hits the curve at the point x=2a. Any help?! PLEASE!

The slope of a curve is at the point (x,y) is 4x3. Find the curve if it is required to pass through the point (1,1). Work... 4(1)3=1 y1=1(x1) y=x

A curve has implicit equation x^22xy+4y^2=12 a)find the expression for dy/dx in terms of y and x. hence determine the coordinates of the point where the tangents to the curve are parallel to the xaxis. b)Find the equation of the normal to the curve at

Consider the curve defined by 2y^3+6X^2(y) 12x^2 +6y=1 . a. Show that dy/dx= (4x2xy)/(x^2+y^2+1) b. Write an equation of each horizontal tangent line to the curve. c. The line through the origin with slope 1 is tangent to the curve at point P. Find the

Suppose that f(x) is an invertible function (that is, has an inverse function), and that the slope of the tangent line to the curve y = f(x) at the point (2, –4) is –0.2. Then: (Points : 1) A) The slope of the tangent line to the curve y = f –1(x) at


Linear approximation: Consider the curve defined by 8x^2 + 5xy + y^3 = 149 a. find dy/dx b. write an equation for the tangent line to the curve at the point (4,1) c. There is a number k so that the point (4.2,k) is on the curve. Using the tangent line

Could someone please help me with these tangent line problems? 1) Find the equation of the line tangent to the given curve at the indicated point: 3y^3 + 2x^2 = 5 at a point in the first quadrant where y=1. 2) Show that there is no point on the graph of

Consider the curve given by the equation y^3+3x^2y+13=0 a.find dy/dx b. Write an equation for the line tangent to the curve at the point (2,1) c. Find the minimum ycoordinate of any point on the curve. the work for these would be appreciated i don't need

1. A solution is prepared such that it is 0.45 M in formic acid and 0.35 M in sodium formate. a) Where is this mixture located on a titration curve: before the buffer point, at the buffer point, or after the buffer point? b) Use the HendersonHasselbach to

Write as a single logarithm: 2 log 3 – log 5 +2 log2

What is the value of the zscores if the area of the curve to this point is 0.887? How much of this are is between the peak of the curve and this point?

Find the inverse of each relation: y = (0.5)^(x+2) and y = 3log base 2 (x3) + 2 For the first one I got y=log base 0.5 (x+2)...but the answer in the back of the textbook says that it is not x+2, but x2. Can someone tell me why it would end up being x2

Please check answers: Use the equation of the exponential function whose graph passes through the points (0,2) and (2,50) to find the value of y when x= 2. My answer: 2/25 Solve 64^x<32^x+2 Answer: x<10 Write the equation log(243)81=4/5 Answer:

Consider line segments which are tangent to a point on the right half (x>0) of the curve y = x^2+7 and connect the tangent point to the xaxis. If the tangent point is close to the yaxis, the line segment is long. If the tangent point is far from the

solve log(2+x)log(x3)=log2 I know the answer is 8


Consider the paraboloid z=x^2+y^2. The plane 3x2y+z7=0 cuts the paraboloid, its intersection being a curve. What is the "the natural" parametrization of this curve? Hint: The curve which is cut lies above a circle in the xyplane which you should

Consider the paraboloid z=x^2+y^2. The plane 3x2y+z7=0 cuts the paraboloid, its intersection being a curve. What is the "the natural" parametrization of this curve? Hint: The curve which is cut lies above a circle in the xyplane which you should

The line that is normal to the curve x^2=2xy3y^2=0 at(1,1) intersects the curve at what other point? Please help. Thanks in advance. We have x2=2xy  3y2 = 0 Are there supposed to be 2 equal signs in this expression or is it x2 + 2xy  3y2 = 0 ? I'll

log 3x=log2+log(x+5)

solve log(2+x)log(x3)=log2

I'm desperate! Find the point where the curve r(t)=(12sint)i  12(cost)j+ 5tk is at a distance 13pi units along the curve from the point (0,12,0) in the direction opposite to the direction of increasing arc length. Thanks for any advice...

there are two tangents lines to the curve f(x) = 3x^2 that pass through the point p =0,1 find the x coordinates of the point where the tangents line intersect the curve

Sorry but I've got a lot of problems that I don't understand. 1) Let f(x)= (3x1)e^x. For which value of x is the slope of the tangent line to f positive? Negative? Zero? 2) Find an equation of the tangent line to the oven curve at the specified point.

Prove that log a, log ar, log ar^2 is an a.p. Is the following below correct? Log ar^2  Log ar= Log ar  Log a hence applying laws of logarithm Log(ar^2/ar) = log(ar/a) Log and log cancels out and then crossmultiply hence a^2r^2 = a^2r^2 L.H.S=R.H.S

I'm studying for my precalc exam and have completed 42 practice problems. I have 3 I need to answer that I can't. Please help. What is the base of the function G(x) = log subscript b x if it's graph has points (16,4)? Using the properties of logarithms how


A(1,0) is a point on the parabola y=2x(x−1). From point A, point P is moving along the curve towards the origin O(0,0). As P → O, sec^2∠APO → N, where N is a positive integer. What is the value of N?

(1,0) is a point on the parabola y=2x(x−1). From point A, point P is moving along the curve towards the origin O(0,0). As P→O, sec^2∠APO→N, where N is a positive integer. What is the value of N?

A point is moving along the curve xy=12. When the point is at (4,3), the xcoordinate decreases at the rate of 2cm/sec. How fast is the ycoordinate changing at that point?

You did just fine and your second derivative is correct, if you meant (6y^2  4x^2)/(9y^3) except they took it a bit further. notice your numerator is 4x^2 + 6y^2 from the original 2x^2  3y^2 = 4 , then 4x^2  6y^2 = 8 , and 4x^2 + 6y^2 = 8 to get

You did just fine and your second derivative is correct, if you meant (6y^2  4x^2)/(9y^3) except they took it a bit further. notice your numerator is 4x^2 + 6y^2 from the original 2x^2  3y^2 = 4 , then 4x^2  6y^2 = 8 , and 4x^2 + 6y^2 = 8 to get