# Multiplication and Subtraction of logs with different bases.

• May 30th 2014, 03:58 AM
NameIsHidden
Multiplication and Subtraction of logs with different bases.
$\text{If }a=\log_{12} 18, b=\log_{24}54\text{, then find the value of }ab+5(a-b)$

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 AM
Plato
Re: Multiplication and Subtraction of logs with different bases.
Quote:

Originally Posted by NameIsHidden
$\text{If }a=\log_{12} 18, b=\log_{24}54\text{, then find the value of }ab+5(a-b)$
I am trying in different ways to equalize the bases. But I can't do anything as the bases are unequal.

Write as $a=\dfrac{\log(2)+2\log(3)}{2\log(2)+\log(3)}$ do the same for $b$.
• May 30th 2014, 05:37 AM
HallsofIvy
Re: Multiplication and Subtraction of logs with different bases.
Quote:

Originally Posted by Plato
Write as $a=\dfrac{\log(2)+2\log(3)}{2\log(2)+\log(3)}$ do the same for $b$.

where the "log" on the right can be to any one base. Probably simplest to use base 10.
• May 30th 2014, 06:01 AM
NameIsHidden
Re: 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 AM
Soroban
Re: 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:

$\text{If }a\,=\, \log_{12}18,\;b\,=\,\log_{24}54.\;\text{ then find the value of }ab+5(a-b)$

$\begin{array}{cccccccc}a \:=\: \log_{12}18 &\Rightarrow& 12^a \:=\: 18 &\Rightarrow& (2^23)^a \:=\: 18 &\Rightarrow& 2^{2a}3^a \:=\: 18 & [1] \\ \\ b \:=\: \log_{24}54 &\Rightarrow& 24^b \:=\: 54 &\Rightarrow& (2^33)^b \:=\: 54 &\Rightarrow& 2^{3b}3^b \:=\: 54 & [2] \end{array}$

Divide [2] by [1]: . $\frac{2^{3b}3^b}{2^{2a}3^a} \:=\:\frac{54}{18} \quad\Rightarrow\quad 2^{3b-2a}3^{b-a} \:=\:3 \:=\:2^03^1$

Equate exponents: . $\begin{Bmatrix}3b-2a &=& 0 \\ b-a &=& 1 \end{Bmatrix}$

Solve the system: . $\begin{bmatrix}a \:=\:\text{-}3 \\ b \:=\:\text{-}2 \end{bmatrix} \;\; \color{red}{??}$
• May 30th 2014, 07:45 AM
NameIsHidden
Re: 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 AM
Plato
Re: Multiplication and Subtraction of logs with different bases.
Quote:

Originally Posted by NameIsHidden
Is a -3??? No I used a calculator and got 1.163.........

$a=$ This

And $b=$ this
• May 30th 2014, 07:39 PM
NameIsHidden
Re: 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?