This is concerning a question in stack exchange : Sum of real values of $x$ satisfying the equation $(x^2-5x+5)^{x^2+4x-60}=1$. I was actually wondering why the correct result is not obtained when applying $\log$ on both sides,like how the second answer does.Why is this so? Is it possible to get this answer using logarithms?
Why can’t I find the value of $x$ using logarithms
algebra-precalculusexponential functionlogarithmsproof-writing
Best Answer
Yes, you are right! We can use logarithm here: $$(x^2+4x-60)\ln(x^2-5x+5)=0,$$ which gives $$x^2-5x+5=1$$ or $$x^2+4x-60=0.$$ By the way, I think the accepted answer on your link is total wrong because if we wrote $$(x^2-5x+5)^{x^2+4x-60}$$ then $x^2-5x+5>0$ by definition.