Logarithms and exponential functions Watch
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
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.