How do you solve #4^x-2^(x+1)=3#?

1 Answer
A. S. Adikesavan
Apr 27, 2016

#x = log 3/log 2#..

Explanation:

Use #a^(mn)=(a^m)^n=(a^n)^m and a^(m+n)=a^ma^n#

#4^x=(2^2)^x=2^(2x)=(2^x)^2#

The given equation is #u^2-2u-3=0#, where #u = 2^x#
The roots are u = #2^x =3 and -1#. As #2^x>0#, for all # x, -1# is inadmissible..

Now solve #2^x=3#, by equating the logarithms.

#x=log 3/log 2# .

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