How do you solve #4^x= 7#?

1 Answer
Gió
Jul 17, 2015

I found: #x=1.4036#

Explanation:

If you can use a pocket calculator (or tables) you could change it into a log using the definition of logarithm:
#log_ab=x->a^x=b#
where in your case you get:
#log_4(7)=x#
you can now change base of your log to evaluate it using a pocket calculator (for example, using the base #e# of the natural logarithm, #ln#):
#x=log_4(7)=(ln(7))/(ln(4))=1.4036#

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