# 12^a = 18 & 24^b=16 then find the value of b in terms of a ?

$\textcolor{b l u e}{b = \frac{\log 8 + \log {12}^{a} - \log 9}{\log} 24}$

with $a = \log \frac{18}{\log} 12 = 1.163171163$ and $b = \log \frac{16}{\log} 24 = 0.8724171679$

#### Explanation:

From the given equations ${12}^{a} = 18$ and ${24}^{b} = 16$
Solution:
From ${12}^{a} = 18$ , we divide both sides of the equation by $9$

${12}^{a} / 9 = \frac{18}{9}$

${12}^{a} / 9 = 2$first equation

From ${24}^{b} = 16$, we divide both sides of the equation by $8$
${24}^{b} / 8 = \frac{16}{8}$

${24}^{b} / 8 = 2$second equation

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$2 = 2$

${24}^{b} / 8 = {12}^{a} / 9$

Multiply both sides by $8$

${24}^{b} / 8 = {12}^{a} / 9$

${24}^{b} = 8 \cdot {12}^{a} / 9$

Take the logarithm of both sides of the equation

$\log {24}^{b} = \log 8 \cdot {12}^{a} / 9$

$b \cdot \log 24 = \log 8 + \log {12}^{a} - \log 9$

Divide both sides by $\log 24$

$b = \frac{\log 8 + \log {12}^{a} - \log 9}{\log} 24$

God bless....I hope the explanation is useful.