Question

What is the solution to this equation? 2ln4+lnx=3ln2+ln(x+1)

Answer 1

`x=1`

Explanation:

Given equation:

`2 \ln 4+\ln x=3 \ln 2+\ln(x+1)`

`\textsf{Apply the power law}: \quad n \ln x = \ln x^n`

`\implies \ln 4^2+\ln x=\ln 2^3+\ln(x+1)`

Simplify:

`\implies \ln 16+\ln x=\ln 8+\ln(x+1)`

Subtract ln 8 from both sides:

`\implies \ln 16+\ln x- \ln 8=\ln(x+1)`

Subtract ln x from both sides:

`\implies \ln 16-\ln 8=\ln(x+1)-\ln x`

`\textsf{Apply the quotient law}: \quad \ln x - \ln y = \ln (x)/(y)`

`\implies \ln \left(16)/(8)\right = \ln \left((x+1)/(x)\right)`

Simplify:

`\implies \ln 2 = \ln \left((x+1)/(x)\right)`

`\textsf{Apply the equality law}: \quad \textsf{if }\: \ln x= \ln y\:\textsf{ then }\:x=y`

`\implies 2=(x+1)/(x)`

Multiply both sides by x:

`\implies 2x=x+1`

Subtract x from both sides:

`\implies x=1`