Question

ln(y+1)-lny=1+3lnx . Express y in terms of x, in a form not involving logarithms. (4 marks, Maths CIE A level)

Answer 1

$remember$
$ln(x)=log_e(x)$
$and$
$log_x(a)-log_x(b)= log_x( \frac{a}{b} )$
$and$
$log_a(b)=c$ means $a^c=b$
$and$
$blog(a)=log(a^b)$



$so$

$ln(y+1)-ln(y)=1+3ln(x)$
$ln( \frac{y+1}{y} )=1+ln(x^3)$

$minus$ $ln(x^3)$ $from$ $both$ $sides$

$ln( \frac{y+1}{y}-ln(x^3) )=1$
$ln( \frac{y+1}{yx^3})=1$
$log_e( \frac{y+1}{yx^3})=1$

$translate$

$e^1=\frac{y+1}{yx^3}$
$e=\frac{y+1}{yx^3}$

$times$ $both$ $sides$ $by$ $yx^3$

$eyx^3=y+1$

$minus$ $y$ $from$ $both$ $sides$

$eyx^3-y=1$
$y(ex^3-1)=1$

$divide$ $both$ $sides$ $by$ $(ex³-1)$

$y= \frac{1}{ex^3-1}$

Answer 2

Ln(y+1)-Lny=1+3 Lnx
$ln \frac{y+1}{y} =lne+ln x^{3} ln \frac{y+1}{y} =ln e x^3, \frac{y+1}{y} =ex^3, 1+ \frac{1}{y} =e x^3, \frac{1}{y} =ex^3-1,$$y= \frac{1}{e x^{3} -1}$