How do you solve $27^(x-3)=1/9$ ?

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Answer 1 · 2015-11-29T09:41:16

Express as powers of $3$ and hence find $x = 7/3$

If $a > 0$ then $a^(bc) = (a^b)^c$

So:

$3^(3(x-3)) = (3^3)^(x-3) = 27^(x-3) = 1/9 = 3^-2$

Since $f(x) = 3^x$ is a one-one function from $RR->(0, oo)$ , the Real solutions of this must have equal exponents:

$3(x-3) = -2$

Hence:

$3x = -2+9 = 7$

So $x = 7/3$

How do you solve 27^(x-3)=1/927^(x-3)=1/9?