These are some I chose to practice but I have a hard time knowing how to do it. The answers are online but that doesn't help. These deal with logarithmic functions etc. PLEASE HELP! THANKYOU
1.)solve the equation:
3^(x+1)=27^(x+3)
2.) e^x=5
3.) 2^(3x)+9=25
4.) 4^(x+1)=21
5.) log6 (5x+8)=log6 (13x)
write an exponential function y=ab^x whose graph passes through the given points.
1.) (1,6), (2,36)
2.) (2,16), (3,64)
write a power function y=ax^b whose graph passes through the given points.
1.) (2,2), (4,16)
2.) (3,3), (6,12)
1.) A store begins selling a new type of baseball shoe. The table shows the number (y) of pairs sold during week (x). Find a power model for the following data.
week x| 1 | 2 |
pair y| 10 | 80 |
I'll solve for the first five problems. You have to recall some laws of exponents for these.
(1)
3^(x+1) = 27^(x+3)
Note that we can rewrite 27 as power of 3: 27 = 3^3. Thus,
3^(x+1) = (3^3)^(x+3)
You can distribute the '3' in 3^3 to (x+3).
3^(x+1) = 3^(3(x+3))
3^(x+1) = 3^(3x+9)
Now that they have the same base, we can equate their exponents:
x + 1 = 3x + 9
x - 3x = 9 - 1
-2x = 8
x = -4
(2)
e^x = 5
Get the natural logarithm (ln) of both sides:
ln (e^x) = ln (5)
Note that we can rewrite ln (e^x) into x*ln(e). But ln(e) is equal to 1. So,
x * 1 = ln(5)
x = ln(5)
(3)
2^(3x) + 9 = 25
2^(3x) = 25 - 9
2^(3x) = 16
This is almost similar to #1. We rewrite 16 as power of 2: 16 = 2^4. So,
2^(3x) = 2^4
I believe you know what to do next. ;)
(4)
4^(x+1) = 21
Now, this doesn't have an exact answer. What we'll do is get the ln of both sides:
ln (4^(x+1)) = ln (21)
Again, for ln (4^(x+1)), we can put the exponent outside and leave the base inside, so ln (4^(x+1)) = (x+1)*ln(4).
(x+1) * ln(4) = ln(21)
x + 1 = ln(21) / ln(4)
x = [ ln(21) / ln(4) ] - 1
(5)
log6 (5x+8) = log6 (13x)
If that means "base 6", then simply equate the terms inside the log:
5x + 8 = 13x
5x - 13x = -8
-8x = -8
x = 1
hope this helps~ `u`
For the next questions, they are all solved with the same method. So I'll guide you to solving #1, then try solving the others by following these steps.
We need to find an equation in the form y = ab^x that passes through the points (1,6) and (2,36).
What we'll do is substitute each point to the equation, and solve for 'a' and 'b'
For (1,6):
y = ab^x
6 = ab^1
6 = ab
For (2,36):
y = ab^x
36 = ab^2
From the first equation, (6 = ab), we can have an expression for a, such that:
a = 6 / b
We can substitute this into the second equation:
36 = ab^2
36 = (6 / b) * b^2
36 = 6b^2 / b
36 = 6b
b = 36/6
b = 6
Now that we have a value for b, we can substitute this to first equation to get a:
6 = a(6)
a = 6/6
a = 1
Therefore, your equation is
y = 6^x
hope this helps~ `u`
Sure, I can help you with these questions. Let's go through them one by one.
1.) To solve the equation 3^(x+1) = 27^(x+3), we can simplify both sides of the equation first. Recall that 3^3 = 27, so we can rewrite the equation as 3^(x+1) = (3^3)^(x+3). Using the exponent rule (a^(b*c) = (a^b)^c), we can simplify further to 3^(x+1) = 3^(3*(x+3)).
Now, since the bases are the same, we can equate the exponents. Therefore, we have x + 1 = 3(x + 3). Expanding the right side, we get x + 1 = 3x + 9.
Next, let's isolate the x-term. Subtracting x from both sides, we have 1 = 2x + 9. Subtracting 9 from both sides, we have -8 = 2x. Finally, dividing both sides by 2, we have x = -4.
So, the solution to the equation 3^(x+1) = 27^(x+3) is x = -4.
2.) To solve the equation e^x = 5, we need to take the natural logarithm (ln) of both sides. This will help us isolate x. So, take the ln of both sides:
ln(e^x) = ln(5)
Using the logarithmic property ln(e^a) = a, we can simplify to:
x = ln(5)
So, the solution to the equation e^x = 5 is x = ln(5).
3.) To solve the equation 2^(3x) + 9 = 25, we need to isolate the term with the exponent. So, subtract 9 from both sides:
2^(3x) = 16
Now, rewrite 16 as 2^4:
2^(3x) = 2^4
Since the bases are the same, we can equate the exponents:
3x = 4
Solving for x, divide both sides by 3:
x = 4/3
So, the solution to the equation 2^(3x) + 9 = 25 is x = 4/3.
4.) To solve the equation 4^(x+1) = 21, we need to isolate the term with the exponent. So, take the logarithm (log) of both sides using any base you prefer (usually base 10 or base e):
log(4^(x+1)) = log(21)
Using the logarithmic property log(a^b) = b*log(a), we can simplify to:
(x+1)log(4) = log(21)
Now, divide both sides by log(4):
x+1 = log(21)/log(4)
Subtract 1 from both sides to isolate x:
x = log(21)/log(4) - 1
So, the solution to the equation 4^(x+1) = 21 is x = log(21)/log(4) - 1.
5.) To solve the equation log6(5x+8) = log6(13x), we can use the property of logarithms that states if the base of the logarithms is the same, then the arguments must be equal. In this case, since both logarithms have base 6, we can set the arguments equal to each other:
5x + 8 = 13x
Now, solve for x by subtracting 5x from both sides:
8 = 8x
Dividing both sides by 8, we get:
x = 1
So, the solution to the equation log6(5x+8) = log6(13x) is x = 1.
For the next set of questions on writing exponential and power functions, let's take them one by one. Let me know which one you'd like help with first.