# Solve the equation algebraically using logarithms. Round to the nearest hundredth 7^(3x+9)=12

Solve the equation algebraically using logarithms. Round to the nearest hundredth
${7}^{3x+9}=12$
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Any exponential function can be represented in logarithmic form, as logarithmic functions are the inverses of exponential functions. In the same way, all logarithmic functions can be represented as exponential functions.
Given equation is
${7}^{3x+9}=12$
Taking log on both sides,
$\mathrm{log}\left({7}^{3x+9}\right)=\mathrm{log}\left(12\right)\phantom{\rule{0ex}{0ex}}\left(3x+9\right)\mathrm{log}7=\mathrm{log}12\phantom{\rule{0ex}{0ex}}\left(3x+9\right)=\frac{\mathrm{log}12}{\mathrm{log}7}\phantom{\rule{0ex}{0ex}}3x+9=1.277\phantom{\rule{0ex}{0ex}}3x=1.227-9\phantom{\rule{0ex}{0ex}}x=-\frac{7.773}{3}=-2.591\approx -2.59$
The required solution is $x=-2.59$