#4. Express the logarithm in terms of log2M and log2N:
log2 1/MN
#5. Simplify:
log4 3 - log4 48
#6. Solve the equation:
loga x = 3/2loga 9 + loga 2
#7. Solve the equation:
logb (x2 + 7) = 2/3logb 64
#8. Solve the equation:
loga (3x + 5) - loga (x - 5) = loga 8
log2(1/MN) = -log2(MN) = -(log2M + log2N)
log4(3)-log4(48) = log4(3/48) = log4(1/16) = log4(4^-2) = -2
loga(x) = 3/2 loga(9) + loga(2)
loga(x) = loga(27*2)
x = 54
logb(x^2+7) = logb(16)
x^2+7 = 16
x^2 = 9
x = ±3
loga(3x+5)-loga(x-5) = loga(8)
loga((3x+5)/(x-5)) = loga(8)
(3x+5)/(x-5) = 8
x = 9
#4. Express the logarithm in terms of log2M and log2N:
To express log2 (1/MN) in terms of log2 M and log2 N, we can use the logarithmic identity: log(a/b) = log(a) - log(b).
Therefore, log2 (1/MN) = log2 (1) - log2 (MN).
Since log2 (1) = 0, we can simplify the expression to:
log2 (1/MN) = -log2 (MN).
#5. Simplify:
To simplify log4 3 - log4 48, we can use the logarithmic properties. The subtraction of logarithms can be expressed as division of the numbers inside the logarithms.
Therefore, log4 3 - log4 48 = log4 (3/48).
Simplifying the expression further, we find:
log4 (3/48) = log4 (1/16) = log4 (1) - log4 (16) = 0 - 2 = -2.
#6. Solve the equation:
To solve the equation loga x = (3/2)loga 9 + loga 2, we can use the logarithmic properties.
First, we can rewrite (3/2)loga 9 as loga (9^(3/2)) and simplify it to loga (sqrt(9)).
The equation becomes:
loga x = loga (sqrt(9)) + loga 2.
Using the property of addition of logarithms, we can rewrite the right side of the equation as:
loga (sqrt(9) * 2) = loga (2 * 3) = loga 6.
Therefore, the equation simplifies to:
loga x = loga 6.
Since the logarithms are equal, the bases must also be equal:
x = 6.
#7. Solve the equation:
To solve the equation logb (x^2 + 7) = (2/3)logb 64, we can apply the logarithmic properties.
First, we can rewrite (2/3)logb 64 as logb (64^(2/3)) and simplify it to logb (4^2).
The equation becomes:
logb (x^2 + 7) = logb (4^2).
Using the property of equality of logarithms, we can drop the logarithms and solve for x:
x^2 + 7 = 4^2.
Simplifying the equation:
x^2 + 7 = 16.
Subtracting 7 from both sides:
x^2 = 9.
Taking the square root of both sides:
x = ±3.
The equation has two solutions: x = 3 and x = -3.
#8. Solve the equation:
To solve the equation loga (3x + 5) - loga (x - 5) = loga 8, we can use the logarithmic properties.
The equation can be rewritten as:
loga ((3x + 5)/(x - 5)) = loga 8.
Using the property of equality of logarithms, we can drop the logarithms and solve the equation:
(3x + 5)/(x - 5) = 8.
To solve for x, we will multiply both sides of the equation by (x - 5) to eliminate the fraction:
3x + 5 = 8(x - 5).
Expanding the equation:
3x + 5 = 8x - 40.
Bringing all the x terms to one side and the constant terms to the other side:
5 + 40 = 8x - 3x.
45 = 5x.
Dividing both sides by 5:
x = 9.
Therefore, the solution to the equation is x = 9.
To express the logarithm in terms of log2M and log2N for #4, we can use the logarithmic identity:
loga(b/c) = loga(b) - loga(c)
In this case, we have:
log2(1/MN) = log2(1) - log2(MN)
Since log2(1) is 0 for any base, we are left with:
log2(1/MN) = -log2(MN)
Therefore, the logarithm in terms of log2M and log2N is -log2(MN).
To simplify the expression log4 3 - log4 48 for #5, we can use the logarithmic property:
loga(b) - loga(c) = loga(b/c)
Applying this property, we get:
log4 3 - log4 48 = log4 (3/48)
Simplifying further, we have:
log4 (3/48) = log4 (1/16)
Since 1/16 can be written as 4^(-2), we have:
log4 (1/16) = log4 4^(-2)
Using the logarithmic property:
loga(a^b) = b
We can rewrite the expression as:
log4 4^(-2) = -2
Therefore, the simplified expression is -2.
Moving on to #6, we need to solve the equation loga x = 3/2loga 9 + loga 2.
Using the logarithmic property:
loga(b^c) = cloga(b)
We can rewrite the equation as:
loga x = loga 9^(3/2) + loga 2
Simplifying further:
loga x = loga √(9^3) + loga 2
Since √(9^3) is equal to 27, we have:
loga x = loga 27 + loga 2
Combining the logarithms using the property:
loga(b) + loga(c) = loga(b * c)
We get:
loga x = loga (27 * 2)
Simplifying further:
loga x = loga 54
Since the bases are the same, we can equate the arguments:
x = 54
Therefore, the solution to the equation is x = 54.
Moving on to #7, we have the equation logb (x^2 + 7) = 2/3logb 64.
Using the logarithmic property:
logb(c^d) = dlogb(c)
We can rewrite the equation as:
logb (x^2 + 7) = logb 64^(2/3)
Simplifying further:
logb (x^2 + 7) = logb (4^2)
Since 4^2 is equal to 16, we have:
logb (x^2 + 7) = logb 16
Combining the logarithms using the property:
logb(a) - logb(b) = logb(a/b)
We get:
logb (x^2 + 7) = logb (16/1)
Simplifying further:
logb (x^2 + 7) = logb 16
Since the bases are the same, we can equate the arguments:
x^2 + 7 = 16
Solving the equation:
x^2 = 9
Taking the square root of both sides:
x = ±3
Therefore, the solutions to the equation are x = 3 and x = -3.
Moving on to #8, we have the equation loga (3x + 5) - loga (x - 5) = loga 8.
Using the logarithmic property:
loga(b) - loga(c) = loga(b/c)
We can rewrite the equation as:
loga ((3x + 5)/(x - 5)) = loga 8
Setting the arguments equal to each other, we have:
(3x + 5)/(x - 5) = 8
Cross-multiplying and simplifying:
3x + 5 = 8(x - 5)
3x + 5 = 8x - 40
Subtracting 3x from both sides:
5 = 5x - 40
Adding 40 to both sides:
45 = 5x
Dividing by 5:
x = 9
Therefore, the solution to the equation is x = 9.