Question

Solve it with all the steps included.

Answer 1

Refer to attachment............:-)

Answer 2

Question :

solve for x.

$log_{2}(4 {}^{x + 1} + 4 ) . log_{2}(4 {}^{x} + 1 ) = log_{ \frac{1}{2} }( \frac{1}{8} )$

${\purple{\boxed{\large{\bold{Properties \: of \: Logarithms }}}}}$

$log_{a}(a) = 1$

$log_{a}(b {}^{n} ) = n log_{a}(b)$

$log_{b {}^{c} }(a) = \frac{1}{c} log_{b}(a)$

$log_{a}(x y ) = log_{a}(x ) + log_{a}(y)$

$\huge{ \underline{ \underline{ \green{ \sf{ Detailed \: Answer :}}}}}$

First simplify the terms

LHS :

= $log_{2}(4 {}^{x + 1} + 4 ) . log_{2}(4 {}^{x} + 1 )$

$= log_{2}(4(4 {}^{x} + 1) ). log_{2}(4 {}^{x} + 1 )$

$= ( log_{2}(4) + log_{2}(4 {}^{x} + 1 ) ). log_{2}(4 {}^{x} + 1 )$

$= (2 log_{2}(2) + log_{2}(4 {}^{x} + 1) . log_{2}(4 {}^{x} + 1)$

$(2 + log_{2}(4 {}^{x} + 1 ) . log_{2}(4 {}^{x} + 1)$

RHS :

$= log_{ \frac{1}{2} }( \frac{1}{8} )$

$= log_{ \frac{1}{2} }( (\frac{1}{2}) {}^{3} )$

$= 3 log_{ \frac{1}{2} }( \frac{1}{2} )$

$= 3$

→now equate LHS and RHS

$(2 + log_{2}(4 {}^{x} + 1 ) ) .( log_{2}(4 {}^{x} + 1 ) = 3$

let $4 {}^{x + 1} + 4 = t$

Thus ,

$(2 + t).t = 3$

$t {}^{2} + 2t - 3 = 0$

$t {}^{2} + 3t - t - 3 = 0$

$t(t + 3) - 1(t + 3) = 0$

$(t + 3)(t - 1) = 0$

→t = 1 or -3

•Case 1:

if t = 1

$log_{2}(4 {}^{x} + 1 ) = 1$

$2 = 4 {}^{x} + 1$

$4 {}^{x} = 1$

x = 0

• Case 2:

if x= -3

$log_{2}(4 {}^{x} + 1 ) = -3$

$2 {}^{ - 3} = 4 {}^{x} + 1$

$\frac{1}{8} - 1 = 4 {}^{x}$

$4 {}^{x} = - \frac{7}{8}$

which is not possible

Therefore,

${\red{\boxed{\large{\bold{x = 0 </p><p>}}}}}$

or

$\Large\boxed{\green{ x = </p><p>log_{10}(1)}}$