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How do you solve #sqrt7^(6x)= 49^(x-6)#?

1 Answer

A. S. Adikesavan

Apr 14, 2016

#x = -12#

Explanation:

Replace #sqrt7# by #7^(1/2)# and 49 by #7^2.#.

Now, use #(a^m)^n=a^(mn)#.

#7^(3x)=7^(2x-12)#, Compare exponents..

#3x=2x-12#.

#x=-12#.

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