How do you solve ${\log _3}\left( {4x - 5} \right) = 5$?
Answer
586.2k+ views
Hint: We will use some identities and formulas of logarithmic functions and then, use them on the given expression, we will then obtain the required answer that is the value of x.
Complete step-by-step answer:
We are given that we need to solve ${\log _3}\left( {4x - 5} \right) = 5$.
For this, we will first of all use: ${\log _b}a = \dfrac{{\log a}}{{\log b}}$.
On replacing a by 4x – 5 and b by 3, we will then obtain:-
$ \Rightarrow {\log _3}\left( {4x - 5} \right) = \dfrac{{\log \left( {4x - 5} \right)}}{{\log 3}}$
Putting this with the given right hand side, we will then obtain:-
$ \Rightarrow \dfrac{{\log \left( {4x - 5} \right)}}{{\log 3}} = 5$
On taking the log 3 from division in the denominator of left hand side to multiplication in the numerator of right hand side, we will then obtain the following expression:-
$ \Rightarrow \log \left( {4x - 5} \right) = 5 \times \log 3$ …………….(1)
Now, we know that: $\log {a^b} = b \times \log a$
Replacing a by 3 and b by 5, we will then obtain the following expression:-
$ \Rightarrow 5 \times \log 3 = \log {3^5}$
Putting this in equation number (1), we will then obtain the following expression:-
$ \Rightarrow \log \left( {4x - 5} \right) = \log {3^5}$
Now, we can cancel out log from both the sides to obtain the following expression:-
$ \Rightarrow 4x - 5 = {3^5}$ ……………………(2)
We also know that ${3^5} = 3 \times 3 \times 3 \times 3 \times 3$
On solving it, we will then obtain: ${3^5} = 243$
Putting this in equation number (2) to obtain the following expression:-
$ \Rightarrow 4x - 5 = 243$
Taking the 5 from subtraction in the left hand side to addition in the right hand side, we will then obtain the following expression:-
$ \Rightarrow 4x = 5 + 243$
Simplifying the calculation on right hand side by adding the required quantities to obtain the following expression:-
$ \Rightarrow $ 4x = 248
Dividing both sides by 4 to get the final result written as follows:-
$ \Rightarrow $ x = 62
Hence, the value of x is 62.
Note:
The students must note that when we cut off log from both sides, it is due to a fact hidden beside it. Let us understand it:-
We know that if log a = b, then we have $a = {e^b}$.
Therefore, if we are given that log a = log b, then we get: $a = \log {e^b}$
Now we will use the fact that $\log {a^b} = b \times \log a$.
Therefore, we get: $a = b\log e$.
We know that log e = 1.
Therefore, we have a = b and thus we have the required formula which we used in the above solution.
Complete step-by-step answer:
We are given that we need to solve ${\log _3}\left( {4x - 5} \right) = 5$.
For this, we will first of all use: ${\log _b}a = \dfrac{{\log a}}{{\log b}}$.
On replacing a by 4x – 5 and b by 3, we will then obtain:-
$ \Rightarrow {\log _3}\left( {4x - 5} \right) = \dfrac{{\log \left( {4x - 5} \right)}}{{\log 3}}$
Putting this with the given right hand side, we will then obtain:-
$ \Rightarrow \dfrac{{\log \left( {4x - 5} \right)}}{{\log 3}} = 5$
On taking the log 3 from division in the denominator of left hand side to multiplication in the numerator of right hand side, we will then obtain the following expression:-
$ \Rightarrow \log \left( {4x - 5} \right) = 5 \times \log 3$ …………….(1)
Now, we know that: $\log {a^b} = b \times \log a$
Replacing a by 3 and b by 5, we will then obtain the following expression:-
$ \Rightarrow 5 \times \log 3 = \log {3^5}$
Putting this in equation number (1), we will then obtain the following expression:-
$ \Rightarrow \log \left( {4x - 5} \right) = \log {3^5}$
Now, we can cancel out log from both the sides to obtain the following expression:-
$ \Rightarrow 4x - 5 = {3^5}$ ……………………(2)
We also know that ${3^5} = 3 \times 3 \times 3 \times 3 \times 3$
On solving it, we will then obtain: ${3^5} = 243$
Putting this in equation number (2) to obtain the following expression:-
$ \Rightarrow 4x - 5 = 243$
Taking the 5 from subtraction in the left hand side to addition in the right hand side, we will then obtain the following expression:-
$ \Rightarrow 4x = 5 + 243$
Simplifying the calculation on right hand side by adding the required quantities to obtain the following expression:-
$ \Rightarrow $ 4x = 248
Dividing both sides by 4 to get the final result written as follows:-
$ \Rightarrow $ x = 62
Hence, the value of x is 62.
Note:
The students must note that when we cut off log from both sides, it is due to a fact hidden beside it. Let us understand it:-
We know that if log a = b, then we have $a = {e^b}$.
Therefore, if we are given that log a = log b, then we get: $a = \log {e^b}$
Now we will use the fact that $\log {a^b} = b \times \log a$.
Therefore, we get: $a = b\log e$.
We know that log e = 1.
Therefore, we have a = b and thus we have the required formula which we used in the above solution.
Recently Updated Pages
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred

Master Class 11 Social Science: Engaging Questions & Answers for Success

Master Class 11 Physics: Engaging Questions & Answers for Success

Master Class 11 Maths: Engaging Questions & Answers for Success

Master Class 11 Economics: Engaging Questions & Answers for Success

Master Class 11 Computer Science: Engaging Questions & Answers for Success

Trending doubts
One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

There are 720 permutations of the digits 1 2 3 4 5 class 11 maths CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Draw a diagram of a plant cell and label at least eight class 11 biology CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Discuss the various forms of bacteria class 11 biology CBSE

