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Question

Question: Solve the equation \[\text{lo}{{\text{g}}_{3}}\left( 4x-5 \right)=5\]....

Solve the equation log3(4x5)=5\text{lo}{{\text{g}}_{3}}\left( 4x-5 \right)=5.

Explanation

Solution

This question is from the topic of pre-calculus. In this question, we will find out the value of x. In solving this question, we will remove the term log by using the formula of logarithmic function. After that, we will solve the further question of getting the value of x.

Complete step-by-step solution:
Let us solve this question.
In this question, we have asked to find the value of x from the given equation in the question.
The given equation in the question is
log3(4x5)=5{{\log }_{3}}\left( 4x-5 \right)=5
Using the formula of logarithmic function that is: if logab=c{{\log }_{a}}b=c, then we can write b=acb={{a}^{c}}.
So, we can write the equation log3(4x5)=5{{\log }_{3}}\left( 4x-5 \right)=5 as
(4x5)=35\left( 4x-5 \right)={{3}^{5}}
As we know that 35{{3}^{5}} can also be written as 3×3×3×3×33\times 3\times 3\times 3\times 3, so we can write the above equation as
(4x5)=3×3×3×3×3\Rightarrow \left( 4x-5 \right)=3\times 3\times 3\times 3\times 3
The above equation can also be written as
(4x5)=243\Rightarrow \left( 4x-5 \right)=243
4x5=243\Rightarrow 4x-5=243
Taking the number 5 to the right side of the equation, we can write the above equation as
4x=243+5\Rightarrow 4x=243+5
The above equation can also be written as
4x=248\Rightarrow 4x=248
Now, dividing 4 to the both side of equation, we can write the above equation as
4x4=2484\Rightarrow \dfrac{4x}{4}=\dfrac{248}{4}
The above equation can also be written as
x=2484\Rightarrow x=\dfrac{248}{4}
The above equation can also be written as
x=62\Rightarrow x=62
Hence, we have solved the equation log3(4x5)=5{{\log }_{3}}\left( 4x-5 \right)=5 and got the value of x as 62.

Note: We should have a proper knowledge in the topic of pre-calculus to solve this type of question easily. We should know about the logarithmic functions. Remember the following formulas of logarithmic functions:
If logab=c{{\log }_{a}}b=c, then we can write b=acb={{a}^{c}}
loga1=0{{\log }_{a}}1=0
logbc=logblogc\log \dfrac{b}{c}=\log b-\log c
log(b×c)=logb+logc\log \left( b\times c \right)=\log b+\log c
The above basic formulas should be kept remembered. They make our solving process easy.