Solve for x using properties of logarithms: ln(9)+ln(x+2)=2ln(x+2) ?

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Answer 1 · 2018-02-28T14:41:12

$x = 7$ $x=7$

$ln (9) + 1 n (x + 2) = 2 ln (x + 2)$ $ln(9)+1n(x+2)=2ln(x+2)$

Subtract $2 ln (x + 2)$ $2ln(x+2)$ from both sides:

$ln (9) + 1 n (x + 2) - 2 ln (x + 2) = 0$ $ln(9)+1n(x+2)-2ln(x+2)=0$

Simplify:

$ln (9) - 1 n (x + 2) = 0$ $ln(9)-1n(x+2)=0$

$ln (a) - ln (b) = ln (\frac{a}{b})$ $ln(a)-ln(b)=ln(a/b)$

Hence:

$ln (9) - 1 n (x + 2) = 0 \Rightarrow ln (\frac{9}{x + 2}) = 0$ $ln(9)-1n(x+2)=0=>ln(9/(x+2))=0$

Raising the base $e$ $e$ to these powers:

$e^{ln (\frac{9}{x + 2}) = e^{0}}$ $e^(ln(9/(x+2))=e^0$

$\frac{9}{x + 2} = 1$ $9/(x+2)=1$

$x = 7$ $x=7$

Substituting in original equation:

$ln (9) + 1 n ((7) + 2) = 2 ln ((7) + 2)$ $ln(9)+1n((7)+2)=2ln((7)+2)$

$ln (9) + 1 n (9) = 2 ln (9)$ $ln(9)+1n(9)=2ln(9)$

$2 ln (9) = 2 ln (9)$ $2ln(9)=2ln(9)$