Solve {left(2^-1+4^-1+6^-1+8^-1right)}^x=1

$${ \left( { 2 }^{ -1 } + { 4 }^{ -1 } + { 6 }^{ -1 } + { 8 }^{ -1 } \right) }^{ x } =1$$

Solve for x

$$x=0$$

Steps Using Definition of Logarithm

Calculate 2 to the power of -1 and get \frac{1}{2}.

$$\left(\frac{1}{2}+4^{-1}+6^{-1}+8^{-1}\right)^{x}=1$$

Calculate 4 to the power of -1 and get \frac{1}{4}.

$$\left(\frac{1}{2}+\frac{1}{4}+6^{-1}+8^{-1}\right)^{x}=1$$

Add \frac{1}{2} and \frac{1}{4} to get \frac{3}{4}.

$$\left(\frac{3}{4}+6^{-1}+8^{-1}\right)^{x}=1$$

Calculate 6 to the power of -1 and get \frac{1}{6}.

$$\left(\frac{3}{4}+\frac{1}{6}+8^{-1}\right)^{x}=1$$

Add \frac{3}{4} and \frac{1}{6} to get \frac{11}{12}.

$$\left(\frac{11}{12}+8^{-1}\right)^{x}=1$$

Calculate 8 to the power of -1 and get \frac{1}{8}.

$$\left(\frac{11}{12}+\frac{1}{8}\right)^{x}=1$$

Add \frac{11}{12} and \frac{1}{8} to get \frac{25}{24}.

$$\left(\frac{25}{24}\right)^{x}=1$$

Take the logarithm of both sides of the equation.

$$\log(\left(\frac{25}{24}\right)^{x})=\log(1)$$

The logarithm of a number raised to a power is the power times the logarithm of the number.

$$x\log(\frac{25}{24})=\log(1)$$

Divide both sides by \log(\frac{25}{24}).

$$x=\frac{\log(1)}{\log(\frac{25}{24})}$$

By the change-of-base formula \frac{\log(a)}{\log(b)}=\log_{b}\left(a\right).

$$x=\log_{\frac{25}{24}}\left(1\right)$$

Solve for x (complex solution)

$$x=\frac{2\pi n_{1}i}{\ln(\frac{25}{24})}$$

$$n_{1}\in \mathrm{Z}$$