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5^{-5x+x_{2}+6}=1
Use the rules of exponents and logarithms to solve the equation.
\log(5^{-5x+x_{2}+6})=\log(1)
Take the logarithm of both sides of the equation.
\left(-5x+x_{2}+6\right)\log(5)=\log(1)
The logarithm of a number raised to a power is the power times the logarithm of the number.
-5x+x_{2}+6=\frac{\log(1)}{\log(5)}
Divide both sides by \log(5).
-5x+x_{2}+6=\log_{5}\left(1\right)
By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).
-5x=-\left(x_{2}+6\right)
Subtract x_{2}+6 from both sides of the equation.
x=-\frac{x_{2}+6}{-5}
Divide both sides by -5.
5^{x_{2}+6-5x}=1
Use the rules of exponents and logarithms to solve the equation.
\log(5^{x_{2}+6-5x})=\log(1)
Take the logarithm of both sides of the equation.
\left(x_{2}+6-5x\right)\log(5)=\log(1)
The logarithm of a number raised to a power is the power times the logarithm of the number.
x_{2}+6-5x=\frac{\log(1)}{\log(5)}
Divide both sides by \log(5).
x_{2}+6-5x=\log_{5}\left(1\right)
By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).
x_{2}=-\left(6-5x\right)
Subtract -5x+6 from both sides of the equation.