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

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

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Full solution

Q. Solve: log⁡(2x+8)+1log⁡(2x+8)=2\log(2^{x}+8)+\frac{1}{\log(2^{x}+8)}=2
  1. Convert to Logarithmic Form: Let y=log⁡(2x+8) y = \log(2^x + 8) . Then the equation becomes y+1y=2 y + \frac{1}{y} = 2 .\newlineMultiply through by y y to get y2+1=2y y^2 + 1 = 2y .\newlineRearrange to form a quadratic equation: y2−2y+1=0 y^2 - 2y + 1 = 0 .
  2. Solve Quadratic Equation: Solve the quadratic equation y2−2y+1=0 y^2 - 2y + 1 = 0 .\newlineThis factors to (y−1)2=0 (y-1)^2 = 0 .\newlineSo, y=1 y = 1 .
  3. Substitute and Rewrite: Substitute back for y y : log⁡(2x+8)=1 \log(2^x + 8) = 1 .\newlineRewrite in exponential form: 2x+8=10 2^x + 8 = 10 .\newlineSolve for 2x 2^x : 2x=2 2^x = 2 .
  4. Solve Exponential Equation: Solve 2x=2 2^x = 2 .\newlineSince 21=2 2^1 = 2 , x=1 x = 1 .

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