How do you solve the exponential equation $9^{2 x + 1} = 3^{5 x - 1}$ $9^(2x+1)=3^(5x-1)$ ?

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How do you solve the exponential equation $9^{2 x + 1} = 3^{5 x - 1}$ $9^(2x+1)=3^(5x-1)$ ?

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Answer 1 · 2017-03-05T14:10:30

$x = 3$ $x=3$

Explanation:

Express the left side with a base of 3

$\Rightarrow {(3^{2})}^{2 x + 1} = 3^{5 x - 1}$ $rArr(3^2)^(2x+1)=3^(5x-1)$

$\Rightarrow 3^{4 x + 2} = 3^{5 x - 1}$ $rArr3^(4x+2)=3^(5x-1)$

Since the bases are equal, we can equate the exponents.

$\Rightarrow 5 x - 1 = 4 x + 2$ $rArr5x-1=4x+2$

$\Rightarrow x = 2 + 1 = 3$ $rArrx=2+1=3$

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How do you solve the exponential equation $9^{2 x + 1} = 3^{5 x - 1}$ $9^(2x+1)=3^(5x-1)$ ?

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