How do you solve $6^(3m+2)=1$ ?

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Answer 1 · 2016-12-23T15:18:47

$m = -2/3$

$x^0 = 1 ($ when $x != 0$ )

$therefore 6^0 = 1$

(you could also check this by entering $log_6(1)$ into a calculator, to get $0$ .)

this gives the equation $3m+2 = 0$ .

subtract $2$ :

$3m = -2$

divide by $3$ :

$m = -2/3$

How do you solve 6^(3m+2)=16^(3m+2)=1?