# Logarithms and exponential functionsWatch

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1 week ago
#21
(Original post by Shas72)
Thanks and pls can you help me with one more problem.
Solve 5^(x^2)>2^x, giving your ans in logarithms.
So I did x^2 log 5- xlog2 >0
Factorised got x>0 and x>log2/log5
The ans is x<0 and x>log2/log 5
So iam not understanding how it it x<0
First, its worth noting that the answer must be about right as you have an x^2 on the exponent of the 5 so the sign is irrelevant and 2^x will be smaller when x<0 as its (1/2)^(-x).
Algebraically, you have
x(xlog 5- log2) > 0
You must factorise fully.

So either both terms are positive or both are negative.
x>0 and x>log(2)/log(5) you already have
x<0 and x<log(2)/log(5) is the other one. The second condition is irrelevant as its x < 0.4ish, so x<0 does the job for them both. which is your 2nd solution.
Last edited by mqb2766; 1 week ago
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