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2022-10-08

Solve the exponential equation ${32}^{x+3}={4}^{3x+5}$ for x

{-4}

{5}

{}

{-1}

{-4}

{5}

{}

{-1}

Mckenna Friedman

Beginner2022-10-09Added 10 answers

Given enponential equation

${32}^{x+3}={4}^{3x+5}$

we have to solve it for x. Therefore,

$\Rightarrow {32}^{x+3}={4}^{3x+5}$

on taking ${\mathrm{log}}_{2}$ (Base 2) both sides, we get,

$\Rightarrow (x+3){\mathrm{log}}_{2}32=(3x+5){\mathrm{log}}_{2}4$

$\Rightarrow (x+3){\mathrm{log}}_{2}(2{)}^{5}=(3x+5){\mathrm{log}}_{2}(2{)}^{2}$

$\Rightarrow 5(x+3){\mathrm{log}}_{2}2=2(3x+5){\mathrm{log}}_{2}2$

Since ${\mathrm{log}}_{2}2=1$, so

$\Rightarrow 5(x+3)=2(3x+5)$

$\Rightarrow 5x+15=6x+10$

$\Rightarrow 6x-5x=15-10$

$\Rightarrow 6x-5x=15-10$

$\Rightarrow x=5$

Thus,

x=5 Answer

${32}^{x+3}={4}^{3x+5}$

we have to solve it for x. Therefore,

$\Rightarrow {32}^{x+3}={4}^{3x+5}$

on taking ${\mathrm{log}}_{2}$ (Base 2) both sides, we get,

$\Rightarrow (x+3){\mathrm{log}}_{2}32=(3x+5){\mathrm{log}}_{2}4$

$\Rightarrow (x+3){\mathrm{log}}_{2}(2{)}^{5}=(3x+5){\mathrm{log}}_{2}(2{)}^{2}$

$\Rightarrow 5(x+3){\mathrm{log}}_{2}2=2(3x+5){\mathrm{log}}_{2}2$

Since ${\mathrm{log}}_{2}2=1$, so

$\Rightarrow 5(x+3)=2(3x+5)$

$\Rightarrow 5x+15=6x+10$

$\Rightarrow 6x-5x=15-10$

$\Rightarrow 6x-5x=15-10$

$\Rightarrow x=5$

Thus,

x=5 Answer

Find the local maximum and minimum values and saddle points of the function. If you have three-dimensional graphing software, graph the function with a domain and viewpoint that reveal all the important aspects of the function

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