How do you solve $3^(2x-1) = 729/9^(x+1)$ ?

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Answer 1 · 2015-10-22T10:36:47

The solution is $x=5/4$ .

First of all, note that all numbers involved are powers of three:

So, we can rewrite the equations in terms of powers of three only:

$3^{2x-1} = 3^6/((3^2)^{x+1})$

Now use the rule for power of powers: $(a^b)^c=a^{bc}$ .

$3^{2x-1} = 3^6/(3^{2x+2)$

Now use the rule for dividing powers of a same base: $a^b / a^c = a^{b-c}$ :

$3^{2x-1} = 3^{6-(2x+2)}$

We're finally in the form $3^a=3^b$ , which is true if and only if $a=b$ , so we must solve

$2x-1 = 6-2x-2$ , which we easily rearrange into

$4x=5$ , and solve for $x$ finding $x=5/4$

How do you solve 3^(2x-1) = 729/9^(x+1)3^(2x-1) = 729/9^(x+1)?