How do you solve $9^{x - 1} = 5^{2 x}$ $9^(x-1)= 5^(2x)$ ?

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Answer 1 · 2018-03-02T08:58:45

$x = - 2.150$ $x=-2.150$

Start by taking the logarithm of both sides, in order to bring the exponents down.

${log}_{10} (9^{x - 1}) = {log}_{10} (5^{2 x})$ $log_10(9^(x-1))=log_10(5^(2x))$

One of the log rules is this:

${log}_{a} (x^{n}) = n {log}_{a} (x)$ $log_a(x^n)=nlog_a(x)$

So:

$(x - 1) {log}_{10} (9) = 2 x {log}_{10} (5)$ $(x-1)log_10(9)=2xlog_10(5)$

Evaluate the logarithm and simplify:

$(x - 1) (0.9542) = 2 x (0.6990)$ $(x-1)(0.9542)=2x(0.6990)$

$0.9542 x - 0.9542 = 1.398 x$ $0.9542x-0.9542=1.398x$

$- 0.9542 = 0.4438 x$ $-0.9542=0.4438x$

$x = - 2.150$ $x=-2.150$

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