Solveeit Logo
Login

Question

Question: How do you solve \[\log \left( {\dfrac{1}{{100}}} \right) = \log ({10^{x + 2}})\] ?...

How do you solve log⁡(1100)=log⁡(10x+2)\log \left( {\dfrac{1}{{100}}} \right) = \log ({10^{x + 2}}) ?

Explanation

Solution

Hint : We can solve this using the rules or laws of logarithms. If we have log⁡(x)=log⁡(y)\log (x) = \log (y) we can cancel the logarithmic function we get ⇒x=y \Rightarrow x = y . We have the logarithm rule log⁡xa=alog⁡x\log {x^a} = a\log x . This is also called power rule. The logarithm of an exponential number is the exponent times the logarithm of the base. Using this we can solve this. We need to find the value of ‘x’.

Complete step-by-step answer :
We have log⁡(1100)=log⁡(10x+2)\log \left( {\dfrac{1}{{100}}} \right) = \log ({10^{x + 2}}) .
We know that if log⁡(x)=log⁡(y)\log (x) = \log (y) then x=yx = y . Comparing this we have x=1100x = \dfrac{1}{{100}} and y=10x+2y = {10^{x + 2}} .
Then we have,
⇒1100=10x+2\Rightarrow \dfrac{1}{{100}} = {10^{x + 2}}
⇒1102=10x+2\Rightarrow \dfrac{1}{{{{10}^2}}} = {10^{x + 2}}
It can be written as
⇒10−2=10x+2\Rightarrow {10^{ - 2}} = {10^{x + 2}}
Applying logarithm on both sides we have,
⇒log⁡(10−2)=log⁡(10x+2)\Rightarrow \log ({10^{ - 2}}) = \log ({10^{x + 2}})
We know log⁡xa=alog⁡x\log {x^a} = a\log x , then above becomes:
⇒−2log⁡(10)=(x+2)log⁡(10)\Rightarrow - 2\log (10) = (x + 2)\log (10)
We know that the value of log⁡(10)\log (10) is 1.
⇒−2=(x+2)\Rightarrow - 2 = (x + 2)
Rearranging the above equation we have,
⇒x+2=−2\Rightarrow x + 2 = - 2
Subtracting 2 on both sides we have,
⇒x=−2−2\Rightarrow x = - 2 - 2
⇒x=−4\Rightarrow x = - 4 . Is the required answer.
(Here we note that the base is 10.)
So, the correct answer is “ x = - 4”.

Note : To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is log⁡(x.y)=log⁡(x)+log⁡(y)\log (x.y) = \log (x) + \log (y) . Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. that is log⁡(xy)=log⁡x−log⁡y\log \left( {\dfrac{x}{y}} \right) = \log x - \log y . Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is log⁡xa=alog⁡x\log {x^a} = a\log x . These are the basic rules we use while solving a problem that involves logarithm function.