logarithm equation I need help getting started?

You have to use all the properties of logs to solve. I will write 'log' for 'log base b', since everything is given for the same base. I have to interpret the left side of the equation as log(x). Otherwise, I don't know what to do with 'b^x'. So:

log x = log 8^(2/3) + log 9^(1/2) - log 6 [Using exponential property of logs]

Since 8^(2/3) = 4 and 9^(1/2) = 3 {review fractional exponents]....

log x = log 4 + log 3 - log 6

log x = log (4(3)/6)

log x = log 2

Since log function is 1 to 1, x = 2.

Log base b^x=2/3log base b 8 + 1/2 log base b 9 - log base b 6

That ^ is messing me up

I am going to assume it is NOT a base of b^x

But

Log base b(x)=(2/3)log base b(8) + (1/2) log base b (9) - log base b (6)

Log base b(x)=log base b(8)^(2/3) + log base b (9)^(1/2) - log base b (6)

8^(2/3)=[cuberoot(8)]^2=2^2=4.................9^(1/2)=sqrt9=3

Log base b(x)=log base b(4) + log base b (3) - log base b (6)

Log base b(x)=log base b(4x3/6)

Log base b(x)=log base b(2)

Take the Antilog(Drop the logs

x=2

log_b(x) = (2/3)log_b(8) + (1/2)log_b(9) - log_b(6)

Change the coefficients into powers inside the logs:

log_b(x) = log_b(8^(2/3)) + log_b(9^(1/2)) + log_b(6^(-1))

You can actually simplify this without irrational numbers, so let's do that now:

log_b(x) = log_b(4) + log_b(3) + log_b(1/6)

Combine the logs by multiplying their contents (since they are all separated by addition and have the same base:

log_b(x) = log_b(4 * 3 * 1/6)

Simplify:

log_b(x) = log_b(2)

Perform the inverse operation to cancel the logs (b^___)

b^(log_b(x)) = b^(log_b(2))

x = 2

log2 (x+a million)^((x+a million)^3)=8 (x+a million)^3log2(x+a million)=8 x=a million is a answer. enable's observe f the function defined with the aid of f(x)=(x+a million)^3log2(x+a million)-8 f is defined for x>-a million. The function c defined with the aid of c(x)=(x+a million)^3 is increasing. The function d defined with the aid of d(x)=log2(x+a million) is increasing. The function g defined with the aid of g(x)=(x+a million)^3log2(x+a million) is increasing as a produced from 2 applications that are increasing. f is increasing. f is non-supply up. For -a million<x<a million f(x)<0 f(a million)=0 For x>a million f(x)>0 The equation has have been given an unique answer. x=a million