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

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.