Solve Log Values Problem: Find A and B for \frac{Log A}{Log B} = \frac{2}{3}

[yoleven]

Homework Statement

A professor is doing a problem on the black board and ends up with the expression

\frac{Log A}{Log B} = \frac{2}{3}
.

He absentmindedly cancels the “log”, making the left-hand side A/B. (A very wrong thing to do!)
Luckily, he ends up with the correct values for A and B. What are these values?

The Attempt at a Solution

I wasn't sure what to do here. I figured 2/3 is .667

from the question I have Log A- Log B= .667
I set A to be equal to 10 which gives me a value of 1, then Log B would equal .333 because 1-.667 =.333

raising both sides to be powers of ten, I get B= 10^.333
which is 2.15

This works except if the proffessor "cancelled the Log) Then 10/2.15 obviously isn't 2/3.
What should I do here?

 

[Office_Shredder]

You can't just set A to be equal to 10. You have two equations

\frac{logA}{logB}=\frac{2}{3} is one of them. Also, \frac{A}{B}=\frac{2}{3} is another. You have to use both of them to solve for A and B

 

[Mark44]

You have an error in your work: (log A)/(log B) != log A - log B. What is true is the log(A/B) = log A - log B.

There are two equations to work with:
A/B = 2/3, and
(log A)/(log B) = 2/3

Solve for one variable in the first equation, and substitute for it in the second equation.

 

[yoleven]

Okay. Here is what I have..

\frac{log A}{log B} = \frac{2}{3}

3 log A = 2 log B

log A³ =log B²

log A³- log B² = 0

log \frac{A^3}{B^2} = 0

\frac{A^3}{B^2} = 1

A³ = B²

from the question, we know that \frac{A}{B} = \frac{2}{3}

B = \frac{3}{2} A

(\frac{3}{2}A)² =A³

\frac{9}{4} A² =A³

A= \frac{9}{4}

B= \frac{3}{2} x \frac{9}{4} = \frac{27}{8}

Thanks for your input.