How do you solve #3 ^(5x) = 27^(2x + 1)#?

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Answer 1 · 2016-04-14T03:46:09

Rearrange and take the logarithms to the base 3 of both sides, to yield #x=-3#

Rewrite the right hand side:

#27^(2x+1)=3^(3(2x+1))=3^(6x+3)#

Now the overall equation is:

#3^(5x)=3^(6x+3)#

Taking the log to the base 3 of both sides:

#5x=6x+3#

#x=-3#