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

Applying logarithm properties, it is found that the correct expression is:

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

These following properties are applied:

$\ln{a} + \ln{b} = \ln{(ab)}$

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

$a\ln{x} = \ln{x^a}$

Also, since $e^1 = e, 1 = \ln{e}$

The original expression is:

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

Since $1 = \ln{e}$

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

Applying the properties, we have that:

$\ln{\left(\frac{y+1}{y}\right)} = \ln{e} + \ln{x^3}$

$\ln{\left(\frac{y+1}{y}\right)} = \ln{ex^3}$

Removing the logarithms

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

$y + 1 = yex^3$

$yex^3 - y = 1$

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

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

A similar problem is given at https://brainly.com/question/14405084