How do you solve $7^{2 x - 1} = 343$ $7^(2x-1) = 343$ ?

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Answer 1 · 2016-03-12T02:02:51

$x = 2$ $x=2$

We will use the following properties of logarithms:

$7^{2 x - 1} = 343 = 7^{3}$ $7^(2x-1) = 343=7^3$

${log}_{7} (7^{2 x - 1}) = {log}_{7} (7^{3})$ $log_7(7^(2x-1))=log_7(7^3)$

$(2 x - 1) {log}_{7} (7) = 3 {log}_{7} (7)$ $(2x-1)log_7(7) = 3log_7(7)$

$2 x - 1 = 3$ $2x-1 = 3$

$2 x = 4$ $2x = 4$

$x = 2$ $x = 2$

How do you solve 72x−1=3437^(2x-1) = 343?