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How do you solve the exponential equation $7^{x} = 7^{2 x - 3}$ $7^x=7^(2x-3)$ ?

1 Answer

Feb 17, 2017

$x = 3$ $x=3$

Explanation:

$7^{x} = 7^{2 x - 3}$ $7^x=7^(2x-3)$

$x log 7 = (2 x - 3) log 7$ $xlog7=(2x-3)log7$

$x = 2 x - 3$ $x=2x-3$

$x = 3$ $x=3$

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