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#