Question

Solve the following equation:
\[{{\log }_{4}}({{2.4}^{x-2}}-1)+4=2x\]
A). \[x=16\]
B). \[x=8\]
C). \[x=3\]
D). \[x=2\]

Answer 1

Hint: Here, to solve this problem we need to first simplify the equation by adding \[-4\] on both sides. After that we have to use the property of logarithms as \[a^{{{\log }_{a}}x}={{a}^{x}}\]. Then we can simplify and take antilog on both sides which will finally give us an equation with a single variable. Solving this we will get our answer as \[x=2\].

Complete step-by-step solution:
We have the equation that is \[{{\log }_{4}}({{2.4}^{x-2}}-1)+4=2x---(1)\]
To solve this type of question,
First of all, we need to add \[-4\] on both sides on equation (1) we get:
\[\Rightarrow {{\log }_{4}}({{2.4}^{x-2}}-1)+4-4=2x-4\]
Here, \[+4\] and \[-4\] get cancelled on LHS we get:
\[\Rightarrow {{\log }_{4}}({{2.4}^{x-2}}-1)=2x-4\]
We know that the antilogarithmic of \[{{\log }_{a}}x={{a}^{x}}\]. So, we can take the antilogarithmic on both sides we get:
\[\Rightarrow {{4}^{{{\log }_{4}}({{2.4}^{x-2}}-1)}}={{4}^{2x-4}}\]
We know that \[{{a}^{{{\log }_{a}}x}}=x\]
\[\Rightarrow {{2.4}^{x-2}}-1={{4}^{2x-4}}\]
By rearranging the term, we get:
\[\Rightarrow {{\left( {{4}^{x-2}} \right)}^{2}}+{{2.4}^{x-2}}-1=0\]
By using the basic property of mathematics that is \[{{(a-b)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab\], we get:
\[\Rightarrow {{\left( {{4}^{x-2}}-1 \right)}^{2}}=0\]
By squaring on both sides, we get:
\[\Rightarrow {{4}^{x-2}}-1=0\]
By adding 1 on both sides, we get:
\[\Rightarrow {{4}^{x-2}}-1+1=0+1\]
By simplifying this we get:
\[\Rightarrow {{4}^{x-2}}=1\]
We can also write 1 on RHS as \[{{4}^{0}}\] because, \[{{4}^{0}}=1\] by applying this on above equation we get:
\[\Rightarrow {{4}^{x-2}}={{4}^{0}}\]
By comparing on both sides, we get:
\[\Rightarrow x-2=0\]
By further solving this above equation we get:
\[x=2\]
So, the correct option is “option D”.

Note: The concept of the logarithm is used to solve this problem. By properties of logarithms, \[a^{{{\log }_{a}}x}={{a}^{x}}\]. Complex multiplication and division are done using logarithm characteristics. To acquire the needed answer, we take the logarithm of the expression, do the operations, and then take the antilog. When a number is expressed as an exponent of a, the logarithm to the base a can be defined as the power of a. The inverse operation of the logarithm is the antilogarithm or exponent.