Solve: log(2^(x)+8)+(1)/(log(2^(x)+8))=2

Convert to Logarithmic Form: Let $ y = \log(2^x + 8) $. Then the equation becomes $ y + \frac{1}{y} = 2 $. Multiply through by $ y $ to get $ y^2 + 1 = 2y $. Rearrange to form a quadratic equation: $ y^2 - 2y + 1 = 0 $.Solve Quadratic Equation: Solve the quadratic equation $ y^2 - 2y + 1 = 0 $. This factors to $ (y-1)^2 = 0 $. So, $ y = 1 $.Substitute and Rewrite: Substitute back for $ y $: $ \log(2^x + 8) = 1 $. Rewrite in exponential form: $ 2^x + 8 = 10 $. Solve for $ 2^x $: $ 2^x = 2 $.Solve Exponential Equation: Solve $ 2^x = 2 $. Since $ 2^1 = 2 $, $ x = 1 $.

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Solve: log(2^(x)+8)+(1)/(log(2^(x)+8))=2

Solve: $\log(2^{x}+8)+\frac{1}{\log(2^{x}+8)}=2$

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Math Problems

Algebra 2

Sum of finite series not start from 1

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Q. Solve:

\log(2^{x}+8)+\frac{1}{\log(2^{x}+8)}=2

Convert to Logarithmic Form: Let $y = \log(2^x + 8)$ . Then the equation becomes $y + \frac{1}{y} = 2$ . $\newline$ Multiply through by $y$ to get $y^2 + 1 = 2y$ . $\newline$ Rearrange to form a quadratic equation: $y^2 - 2y + 1 = 0$ .
Solve Quadratic Equation: Solve the quadratic equation $y^2 - 2y + 1 = 0$ . $\newline$ This factors to $(y-1)^2 = 0$ . $\newline$ So, $y = 1$ .
Substitute and Rewrite: Substitute back for $y$ : $\log(2^x + 8) = 1$ . $\newline$ Rewrite in exponential form: $2^x + 8 = 10$ . $\newline$ Solve for $2^x$ : $2^x = 2$ .
Solve Exponential Equation: Solve $2^x = 2$ . $\newline$ Since $2^1 = 2$ , $x = 1$ .