Question

Solve for x. Log5 (x+3)=1-log5 (x-1).

Answer 1

${\huge {\mathfrak {Answer :-}}}$

➡️ $log_5 (x + 3) = 1 - log_5 (x - 1)$

Or, $log_5 (x + 3) = log_5 5 - log_5 (x - 1)$

Or, x + 3 = 5 / (x - 1)

Or, x^2 - x + 3x - 3 = 5

Or, x^2 + 2x - 8 = 0

Or, x^2 + 4x - 2x - 8 = 0

Or, x(x + 4) - 2(x + 4) = 0.

Or, (x - 2)(x + 4) = 0

Or, x - 2 = 0.

Or, x = 2.

➡️ Hence, x = 2.

That's it..

Answer 2

$\huge\bf\mathscr\pink{Your\: Answer}$

x = 2

step-by-step explanation:

Given,

$log_5(x+3)=1-log_5(x-1)$

Now,

we know that,

$log_aa$ cm = 1

so,

we may write,

1 = $log_55$

now,

putting this value in the Equation,

we get,

$log_5(x+3)=log_55-log_5(x-1)$

=> $log_5(x+3)= log_5(5/x-1)$

{ °.° log a - log b = log (a/b) }

Now,

all the quantities are on the same base '5'

so,

we can simpliy write,

=> x+3 = 5/(x-1)

doing cross multiply,

=> (x-1)(x+3) = 5

=> ${x}^{2}+2x-3-5=0$

=> ${x}^{2}+2x-8=0$

=> ${x}^{2}+4x-2x-8=0$

Doing factorisation,

we get,

=> $x(x+4)-2(x+4)=0$

=> $(x-2)(x+4)=0$

=> x-2 = 0 and x+ 4=0

=> x = 2 and x = -4

But,

if we put

x = -4

then,

$log_5(x+3)$

= $log_5(-4+3)$

= $log_5(-1)$

But,

we know that,

in $log_ab$,

b > 0

therefore,

it is contradiction,

Hence,

x = 2