How do you solve \[ {2^x} = 27 \\ \\ \]?
Answer
573.9k+ views
Hint:Whenever there comes a question in which the variable is placed in the exponent then always take the natural logarithm of both the sides of the equation so that the variable placed as the exponent can be removed and solved further.
Complete Step by Step Solution:
When we take the natural log we write as $
\ln ({2^x}) \\ \\
$ or ${\log _e}({2^x})$ with its base as mathematical constant e, where e is approximately equal to 2.71 , sometimes we generally write $\log ({2^x})$ as we assume base implicitly written there with logarithm.
For now, we can solve the question as :
$\ln ({2^x}) = \ln (27)$
By expanding $
\ln ({2^x}) \\
\\
$ by moving x in front of logarithm as per the rule of logarithm which is stated as ${\log _a}{x^p} =
p{\log _a}x$ , also known as Power Rule is applied on this question. So, the next statement
For our answer will be
$x\ln (2) = \ln (27)$
In order to solve further we will divide both the terms that is L.H.S and R.H.S by $\ln (2)$and by dividing each term we will get
\[
\dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\
\\
\]$
\dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\
\\
$
To Simplify further we need to cancel the common factor of $\ln (2)$.
\[
\dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\
\\
\]
Further dividing $x$ by 1 we finally get the equation as-
\[x = \dfrac{{\ln (27)}}{{\ln (2)}}\]
The result can be calculated in decimal form by using a logarithmic table . Here the answer in Decimal Form would be \[x = 4.75488750\] and concluding at the step \[x = \dfrac{{\ln (27)}}{{\ln (2)}}\] will be the final answer to this question.
Note:Remember the rules of logarithms as they are the key to solve any question related to logarithm . in this question we used Power Rule ${\log _a}{x^p} = p{\log _a}x$ and Whenever there comes a question in which the variable is placed in the exponent then always take the natural logarithm (natural log means having base as ‘e’ which is approximately equal to 2.71).
Complete Step by Step Solution:
When we take the natural log we write as $
\ln ({2^x}) \\ \\
$ or ${\log _e}({2^x})$ with its base as mathematical constant e, where e is approximately equal to 2.71 , sometimes we generally write $\log ({2^x})$ as we assume base implicitly written there with logarithm.
For now, we can solve the question as :
$\ln ({2^x}) = \ln (27)$
By expanding $
\ln ({2^x}) \\
\\
$ by moving x in front of logarithm as per the rule of logarithm which is stated as ${\log _a}{x^p} =
p{\log _a}x$ , also known as Power Rule is applied on this question. So, the next statement
For our answer will be
$x\ln (2) = \ln (27)$
In order to solve further we will divide both the terms that is L.H.S and R.H.S by $\ln (2)$and by dividing each term we will get
\[
\dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\
\\
\]$
\dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\
\\
$
To Simplify further we need to cancel the common factor of $\ln (2)$.
\[
\dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\
\\
\]
Further dividing $x$ by 1 we finally get the equation as-
\[x = \dfrac{{\ln (27)}}{{\ln (2)}}\]
The result can be calculated in decimal form by using a logarithmic table . Here the answer in Decimal Form would be \[x = 4.75488750\] and concluding at the step \[x = \dfrac{{\ln (27)}}{{\ln (2)}}\] will be the final answer to this question.
Note:Remember the rules of logarithms as they are the key to solve any question related to logarithm . in this question we used Power Rule ${\log _a}{x^p} = p{\log _a}x$ and Whenever there comes a question in which the variable is placed in the exponent then always take the natural logarithm (natural log means having base as ‘e’ which is approximately equal to 2.71).
Recently Updated Pages
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

Master Class 11 Chemistry: 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

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State and prove Bernoullis theorem class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

Which among the following are examples of coming together class 11 social science CBSE

