Question

Solve Log4x + log4(x-6)=2

Answer 1

Step-by-step explanation:

log4x+log4(x-6)=2

place under single log using multiplication rule

log4(x(x-6))=2

convert to exponential form:(base(4) raised to log of number(2)=number(x(x-6)

4^2=x(x-6)

16=x^2-6x

x^2-6x-16=0

(x-8)(x+2)=0

x=8

or

x=-2 (reject, x>0)

HOPE IT HELPS ....!

Answer 2

GIVEN :–

$\\ \: \: { \huge{.}} \: \: { \bold{ log_{4}(x) + log_{4}(x - 6) = 2 }} \: \:$

TO FIND :–

• Value of 'x' = ?

SOLUTION :–

$\\ \: \implies \: { \bold{ log_{4}(x) + log_{4}(x - 6) = 2 }} \: \:$

• Using property –

$\\ \: \longrightarrow \: \large { \boxed { \bold{ log_{e}(a) + log_{e}(b) = log_{e}(a.b) }}} \: \:$

• So that –

$\\ \: \implies \: { \bold{ log_{4}(x(x - 6) ) = 2 }} \: \:$

$\\ \: \implies \: { \bold{ log_{4}(x {}^{2} - 6x) = 2 }} \: \:$

• Using property –

$\\ \: \longrightarrow \: \large { \bold { if \: \: log_{e}(a) = b \: \: then \: \: { \boxed{ \bold{ a = {e}^{b}}}} }} \: \:$

• So that –

$\\ \: \implies \: { \bold{ (x {}^{2} - 6x) = {(4)}^{2} }} \: \:$

$\\ \: \implies \: { \bold{ (x {}^{2} - 6x) = 16 }} \: \:$

$\\ \: \implies \: { \bold{ x {}^{2} - 6x - 16 = 0 }} \: \:$

$\\ \: \implies \: { \bold{ x {}^{2} - 8x + 2x- 16 = 0 }} \: \:$

$\\ \: \implies \: { \bold{ x (x- 8) + 2(x- 8) = 0 }} \: \:$

$\\ \: \implies \: { \bold{ (x + 2)(x- 8) = 0 }} \: \:$

$\\ \: \implies \: \large { \boxed{ \bold{ x = - 2 \: \:, \: \: 8}}} \: \:$

• But x > 0 , So that –

$\\ \: \longrightarrow \: \large { \boxed{ \bold{ x = 8}}} \: \:$