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Question

Question: How do you solve \({{\log }_{10}}x=-2\)?...

How do you solve log10x=2{{\log }_{10}}x=-2?

Explanation

Solution

Try to simplify the equation by taking the power of the base on both the sides of the equation. Convert the negative power of base to positive by taking reciprocal. Solve the equation for the value of ‘x’, as per requirement of the question.

Complete step by step answer:
For solving log10x=2{{\log }_{10}}x=-2, we have to find the value of ‘x’ for which it will satisfy the equation.
If ‘log’ of some value ‘m’ with base ‘a’ is ‘n’ i.e. logam=n{{\log }_{a}}m=n, then in order to simplify it we can raise both the sides by the base power.
It can be written as,
logam=n{{\log }_{a}}m=n
Taking power of ‘a’ both the sides, we get
alogam=an\Rightarrow {{a}^{{{\log }_{a}}m}}={{a}^{n}}
For alogam{{a}^{{{\log }_{a}}m}}, both ‘a’ and ‘loga{{\log }_{a}}’ can be cancelled out and simplified as alogam=m{{a}^{{{\log }_{a}}m}}=m
Now, considering our equation log10x=2{{\log }_{10}}x=-2
By raising the power of ‘10’ both the sides, we get
10log10x=102\Rightarrow {{10}^{{{\log }_{10}}x}}={{10}^{-2}}
Here, 10log10x{{10}^{{{\log }_{10}}x}} can be simplified as 10log10x=x{{10}^{{{\log }_{10}}x}}=x
And negative of power ‘n’ of a base ‘m’ i.e. mn{{m}^{-n}} can be made positive by taking reciprocal
So, mn{{m}^{-n}} can be written as 1mn\dfrac{1}{{{m}^{n}}}
x=1102 x=1100 \begin{aligned} & \Rightarrow x=\dfrac{1}{{{10}^{2}}} \\\ & \Rightarrow x=\dfrac{1}{100} \\\ \end{aligned}
This is the solution for the above equation.

Note:
Taking power of base on both sides should be the first approach for solving such questions. Negative to positive base power simplification must be done by taking the reciprocal. Solution should be in maximum simplified form. For example x=1102x=\dfrac{1}{{{10}^{2}}} can also be the solution of the above equation, but it can be further simplified as x=1100x=\dfrac{1}{100}. Which can also be written as x=0.01x=0.01.