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Question: How do you solve \[ {2^x} = 27 \\\ \\\ \]?...

How do you solve {2^x} = 27 \\\ \\\?

Explanation

Solution

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⁡(2x)   \ln ({2^x}) \\\ \\\ or log⁡e(2x){\log _e}({2^x}) with its base as mathematical constant e, where e is approximately equal to 2.71 , sometimes we generally write log⁡(2x)\log ({2^x}) as we assume base implicitly written there with logarithm.
For now, we can solve the question as :
ln⁡(2x)=ln⁡(27)\ln ({2^x}) = \ln (27)
By expanding ln⁡(2x)   \ln ({2^x}) \\\ \\\ by moving x in front of logarithm as per the rule of logarithm which is stated as log⁡axp=plog⁡ax{\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
xln⁡(2)=ln⁡(27)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)\ln (2)and by dividing each term we will get

xln⁡(2)ln⁡(2)=ln⁡(27)ln⁡(2)  \dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\\ \\\

\dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\
\\
ToSimplifyfurtherweneedtocancelthecommonfactorof To Simplify further we need to cancel the common factor of\ln (2)$.

xln⁡(2)ln⁡(2)=ln⁡(27)ln⁡(2)  \dfrac{{x\ln (2)}}{{\ln (2)}} = \dfrac{{\ln (27)}}{{\ln (2)}} \\\ \\\

Further dividing xx by 1 we finally get the equation as-
x=ln⁡(27)ln⁡(2)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=4.75488750x = 4.75488750 and concluding at the step x=ln⁡(27)ln⁡(2)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⁡axp=plog⁡ax{\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).