How do you solve #5^(x+2)=4#?

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Answer 1 · 2016-07-25T15:11:46

# x = (2*(log 2 - log 5))/log 5#

One of the logarithm rules one should keep in mind for this problem:

#log a^b = b*loga#

Apply logarithm on both sides

#log(5^(x+2)) = log 4#

#=>(x+2)*log 5 = log 4#

#=>x+2 = log 4/log 5#

Now it's just a matter of simplification:

#=> x = log(2^2)/log 5 - 2#

#=>x = (2*log 2)/ log 5 - 2#

#=> x = (2*log 2 - 2 log 5)/log 5#

#or, x = (2*(log 2 - log 5))/log 5#