Question

If ${\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = {\log _2}({4^x} + 1) + {\log _2}({4^{x + 1}} + 4)$ then find the value of x will be ?
A. 0
B. 1
C. 2
D. -1

Answer 1

Hint : Here we have to find the value of x. The given question is involving the log terms. we use the properties of logarithmic functions and we simplify the given term. The base value of the log term of LHS is different from the base of the log of the RHS. Hence by the logarithmic properties we determine the value of x.

Complete step by step solution:
Now consider the equation
${\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = {\log _2}({4^x} + 1) + {\log _2}({4^{x + 1}} + 4)$
On simplifying the terms which are present in LHS side
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = {\log _2}{4^x} + {\log _2}1 + {\log _2}{4^{x + 1}} + {\log _2}4$
As we know that ${\log _2}1 = 0$ and the above equation is written as
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = {\log _2}{4^x} + {\log _2}{4^{x + 1}} + {\log _2}{2^2}$
On further simplification the above equation is written as
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = {\log _2}{2^{2x}} + {\log _2}{2^{2(x + 1)}} + {\log _2}{2^2}$
As we know that the property $\log \,{m^n} = n\log m$ , using this property the above equation is written as
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = 2x{\log _2}2 + 2(x + 1){\log _2}2 + 2{\log _2}2$
As we know that the value ${\log _2}2 = 1$ , considering this the above equation is written as
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = 2x + 2(x + 1) + 2$
On further simplification in RHS part
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = 2x + 2x + 2 + 2$
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 8 }} = 4x + 4$
The term which is present in the LHS it can be written as
$\Rightarrow {\log _{\dfrac{1}{{\sqrt 2 }}}}{\left( {\dfrac{1}{{\sqrt 2 }}} \right)^3} = 4x + 4$
As we know that the property $\log \,{m^n} = n\log m$ , using this property the above equation is written as
$\Rightarrow 3{\log _{\dfrac{1}{{\sqrt 2 }}}}\left( {\dfrac{1}{{\sqrt 2 }}} \right) = 4x + 4$
As we know that the value ${\log _{\dfrac{1}{{\sqrt 2 }}}}\dfrac{1}{{\sqrt 2 }} = 1$ , considering this the above equation is written as
$\Rightarrow 3 = 4x + 4$
Take 4 to LHS and we simplify we get
$\Rightarrow 3 - 4 = 4x$
$\Rightarrow - 1 = 4x$
When we divide the above equation by 4 we get
$\Rightarrow x = - 0.25$
When we round off the number we obtain
$\Rightarrow x = 0$
Because the zero is neither positive nor negative
Therefore the option A is the correct one.
Hence we have determined the solution for the given question.
So, the correct answer is “Option A”.

Note : If the question involves the equation and the equation contains the log word then it belongs to the topic logarithm. If the base and argument of a logarithmic function is same then the value will be one. Since the function is logarithm we must know about the properties and laws of logarithmic function.