I am trying in different ways to equalize the bases. But I can't do anything as the bases are unequal.

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- May 30th 2014, 03:58 AMNameIsHiddenMultiplication and Subtraction of logs with different bases.

I am trying in different ways to equalize the bases. But I can't do anything as the bases are unequal. - May 30th 2014, 04:40 AMPlatoRe: Multiplication and Subtraction of logs with different bases.
- May 30th 2014, 05:37 AMHallsofIvyRe: Multiplication and Subtraction of logs with different bases.
- May 30th 2014, 06:01 AMNameIsHiddenRe: Multiplication and Subtraction of logs with different bases.
I got $\frac{6(\log 2)(\log 2)+5(\log 2)(\log 3)+6(\log 3)(\log 3)}{6(\log 2)(\log 2)+5(\log 2)(\log 3)+(\log 3)(\log 3)}$

Is it right saying it $\frac{1}{5(\log 3)(\log 3)}+1$ - May 30th 2014, 06:06 AMSorobanRe: Multiplication and Subtraction of logs with different bases.
Hello, NameIsHidden!

I'm getting incorrect answers and I don't know why.

[I know ... I sound like a kid in Algebra I, but I can't help it.]

Quote:

Divide [2] by [1]: .

Equate exponents: .

Solve the system: .

- May 30th 2014, 07:45 AMNameIsHiddenRe: Multiplication and Subtraction of logs with different bases.
Is a -3??? No I used a calculator and got 1.163.........

- May 30th 2014, 08:33 AMPlatoRe: Multiplication and Subtraction of logs with different bases.
- May 30th 2014, 07:39 PMNameIsHiddenRe: Multiplication and Subtraction of logs with different bases.
I checked with the solution, the result of the expression is $1$.

While taking Soroban's explanation, I see that he missed something.

Actually $a=-3$ or $a=3$ and $b=-2$ or $b=2$.(By substituting two equations in two different ways)

The final result is either $1$ or $11$.

Are there any other ways to compute it even accurate?