Logarithmic problem

Question

If $\log_{16} 15 =a$ and $\log_{12} 18 =b$, then show that $$\log_{25} 24 = \frac{5-b}{16a-8ab-4b+2}.$$

Answer 1

Let $l_2= \ln 2$,$l_3= \ln 3$ and $l_5= \ln 5$. Now change base & we have
\begin{eqnarray*} \frac{5-b}{16a-8ab-4b+2} = \frac{5-\frac{l_2+2l_3}{2l_2+l_3}}{16 \frac{l_3+l_5}{4l_2}-8 \frac{l_3+l_5}{4l_2} \frac{l_2+2l_3}{2l_2+l_3} -4 \frac{l_2+2l_3}{2l_2+l_3} +2} \\ \end{eqnarray*}

\begin{eqnarray*} = \frac{(5(2l_2+l_3)-(l_2+2l_3))l_2} {4 (l_3+l_5)(2l_2+l_3) -2 (l_3+l_5)(l_2+2l_3) -4 (l_2+2l_3)l_2 +2(2l_2+l_3)l_2} \\ \end{eqnarray*}

\begin{eqnarray*} = \frac{(9l_2+3l_3)l_2} {(l_3+l_5)6l_2-6l_2l_3} \\ \end{eqnarray*}

\begin{eqnarray*} = \frac{(3l_2+l_3)} {2 l_5} = \log_{25}(24) \\ \end{eqnarray*}

Answer 2

HINT: your denominator $$16a-8ab-4b+2$$ simplifies to $$6\log_{12} 5$$ and the numerator to $$3\log_{12}24$$

Answer 3

Let $x=\log _{25}24$ so that $25^x=24$

From $16^a=15$ we get $$2^{8a x}=16^{2a x}=15^{2 x}= 3^{2 x}5^{2x}=3^{2x}24=2^33^{2x+1}$$

$$2^{8a x-3}=3^{2x+1}$$

From $12^b=18$ we get

$$2^{2b-1}=3^{2-b}$$

Multiplying these two equations we have

$$(8a x-3)(2-b)=(2b-1)(2x+1)$$

Solve for $x$