If $ \log_{12} 18 = a$, then what is $\log_{24} 16$ equal to?

Question

I was trying to revert the logarithm into indices, but couldn't do it.

There are four options in the answer:

1. $\frac{8-4a}{(5-a)}$
2. $\frac{1}{3+a}$
3. $\frac{4a-1}{2+3a}$
4. $\frac{8-4a}{5+a}$

Answer 1

Here's a solution $$\frac{\log_218}{\log_212}=a$$ $$\frac{2\log_23+1}{\log_23+2}=a$$ which gives $\log_23=\frac{1-2a}{a-2}=\frac{-(1-2a)}{-(a-2)}=\frac{2a-1}{2-a}$ Second is $$\frac{\log_216}{\log_224}$$ $$\frac{4}{\log_23+3}$$ Substitute for $\log_23$ giving...... $$\frac{8-4a}{5-a}$$

Answer 2

Let $\log_32=x$.

Hence, $a=\log_{12}18=\frac{x+2}{1+2x}$, which gives $x=\frac{a-2}{1-2a}$.

Hence, $\log_{24}16=\frac{4x}{3x+1}=\frac{\frac{4(a-2)}{1-2a}}{\frac{3(a-2}{1-2a}+1}=\frac{4(a-2)}{a-5}$.

Answer 3

Using the base-change formula, we can write this as:

$\large\frac{log 18}{log 12} = a \\[0.4cm] \large\frac{2log 3+log 2}{log 3+2log 2}=a\\[0.4cm] 2 log 3 + log 2=a log 3+2alog2\\ (2-a)log3=(2a-1)log2\\[0.4cm] log 3=\frac{(2a-1)log2}{2-a} ...... (1)\\[0.4cm] log_{24}16 = \frac{log24}{log16}\\[0.4cm] \Rightarrow \frac{4log2}{log3+3log2}\\[0.4cm] Now, substituting \text{ } value \text{ } of \text{ } log3: \\ \frac{4log2}{\frac{(2a-1)log2}{2-a}+2log2} = \frac{4(2-a)}{5-a} = \frac{8-4a}{5-a} \text{ } (Ans.)$

Answer 4

Alternatively:

We are given that $ \ \log_{12} 18 \ = \ a \ \ $ or $ \ 12^a \ = \ \ 18 \ \ $ and wish to find $ \ p \ $ such that $ \ 24^p \ = \ 16 \ \ \ . \ $ Starting from the first equation, we have $$ ( \ 2^2·3 \ )^a \ = \ \ 2 · 3^2 \ \ \Rightarrow \ \ 2^{2a \ - \ 1} \ = \ \ 3^{2 \ - \ a} \ \ \Rightarrow \ \ 2^{(2a \ - \ 1)/( 2 \ - \ a)} \ \ = \ \ 3 \ \ . $$

Now that we have a way of "eliminating" factors of $ \ 3 \ \ , \ $ we can insert this into the second equation to obtain $ \ p \ \ : $ $$ ( \ 2^3 · 3 \ )^p \ = \ 2^4 \ \ \Rightarrow \ \ 2^{3p} \ · \ [ \ 2^{(2a \ - \ 1)/( 2 \ - \ a)} \ ]^{ \ p} \ \ = \ \ 2^4 $$ $$ \Rightarrow \ \ 3p \ + \ \frac{p·(2a \ - \ 1)}{ 2 \ - \ a} \ \ = \ \ 4 \ \ \Rightarrow \ \ 3p·(2 \ - \ a) \ + \ p·(2a \ - \ 1) \ \ = \ \ 4·(2 \ - \ a) $$ $$ \Rightarrow \ \ p \ \ = \ \ \frac{4·(2 \ - \ a)}{3 ·(2 \ - \ a) \ + \ (2a \ - \ 1)} \ \ = \ \ \frac{4·(2 \ - \ a)}{6 \ - \ 3a \ + \ 2a \ - \ 1 } \ \ = \ \ \frac{4·(2 \ - \ a)}{5 \ - \ a } \ \ , $$ which is choice $ \ \mathbf{(1)} \ \ . $

As a check, $ \ a \ \approx \ 1.1632 \ \Rightarrow \ p \ \approx \ \frac{4·(2 \ - \ 1.1632)}{5 \ - \ 1.1632 } \ \approx \ 0.8724 \ \ , \ $ which does yield $ \ 24^{0.8724} \ \approx \ 16 \ \ . $

[I've tried some experimentation to see whether if it helpful to manipulate the first equation to produce a factor of $ \ 16 \ $ which could then be replaced by $ \ 24^p \ \ , \ $ but it proves to be better to find a way to "dispose" of the factors of $ \ 2 \ $ or of $ \ 3 \ $ early on.]