Question

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

Answer 1

Answer:

$x=1$

Step-by-step 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:

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

$\textsf{Apply the quotient law}: \quad \ln x - \ln y = \ln \frac{x}{y}$

$\implies \ln \left\dfrac{16}{8}\right = \ln \left(\dfrac{x+1}{x}\right)$

Simplify:

$\implies \ln 2 = \ln \left(\dfrac{x+1}{x}\right)$

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

$\implies 2=\dfrac{x+1}{x}$

Multiply both sides by x:

$\implies 2x=x+1$

Subtract x from both sides:

$\implies x=1$