Question 105757
{{{log(x) + log(x-48) = 2}}}

Read my lesson,"An overview to the laws of logarithms"


{{{log(x(x-48))=2}}}

The antilogarithm of any number a, is the number who has a logarithm of a.


{{{antilog(log(x(x-48)))=antilog(2)}}}


Take note that {{{log(10^2)=2}}}. Thus, {{{antilog(2)=10^2=100}}}

{{{x(x-48)=100}}}
{{{x^2-48x=100}}}
{{{x^2-48x-100=0}}}

By the quadratic formula,

{{{x=(48+-sqrt(48^2-4*1*(-100)))/(2*1)}}}
{{{x=(48+-sqrt(2304+400))/(2)}}}
{{{x=(48+-sqrt(2704))/(2)}}}
{{{x=(48+- 52)/(2)}}}
{{{x=-2}}} or {{{x=50}}}


Power up,
HyperBrain!