sum of alternating base and result

Apply the log rules. $\log_x x = 1$ then if $x=5$ , $\log_5 5 + \log_5 5 = 1 + 1 = 2$ . Log is this. $a^{b} = c$ then $\log_a c =b$ . But log has some rules. If the value of $a$ is negative, then it must be undefinable. And $a$ must not be 0. So $a$ must be a positive number. But if $a$ is $1$ then it is 1. Because $a^{b} = c \Rightarrow 1^{b} = c$ Then $c$ must be 1. Convert this to a logarithm form, $\log_a c = b \Rightarrow \log_1 1 =b$ . Any number of $b$ enters the exponential expression, or log expression, the value must be 1. So $a > 0 , a \neq 1$ . Not end yet. if $c$ is negative, then it is undefinable like $a<0$ . And $c$ should not be 0. Furthermore, $c$ is 1, then it must be 1. Lets summarize it. $a^{b} = c \Rightarrow \log_a c = b$ and $a>0, a \neq 1, c>0$ . And final, $b$ can be a negative number because $a^{b} = c , b<0 \Rightarrow \frac{1}{a^{|b|}}$ .