Why can't I find the value of $x$ using logarithms?

Question

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?

Answer 1

The solution you have mentioned use indeed logarithm, that is for $A>0$

$$A^B=1 \iff \log A^B=\log 1 \iff B\cdot \log A=0 \iff B=0 \quad \lor \quad \log A=0$$

Note that usually $A^B$ is (well-)defined only for $A>0$ but we could also extend the solutions to the case

• $A=-1$
• $B=2k$ with $k\in \mathbb{Z}$

and include also the case $x=2$ among the solutions since $$(-1)^{-52}=\frac1{(-1)^{52}}=1$$

To summarize in order to address your question:

• $A^B=1$ can be solved by logarithm only for the case $A>0$
• $A^B=1$ has also one "special" solution for $A\le 0$ which cannot be find by logarithm but directly by the given conditions

Answer 2

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.