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Question: If we have an expression as \({{\left( 0.2 \right)}^{x}}=2\) and \({{\log }_{10}}2=0.3010\), then wh...

If we have an expression as (0.2)x=2{{\left( 0.2 \right)}^{x}}=2 and log⁡102=0.3010{{\log }_{10}}2=0.3010, then what is the value of xx.

Explanation

Solution

To solve this question we have to take logarithm on both side of equation given (0.2)x=2{{\left( 0.2 \right)}^{x}}=2. Then, by applying the logarithm properties and valve given in the question log⁡102=0.3010{{\log }_{10}}2=0.3010 we obtain a value of xx. The logarithm properties used to solve this question are as following-
log⁡(m)n=nlog⁡m\log {{\left( m \right)}^{n}}=n\log m
log⁡mn=log⁡m−log⁡n\log \dfrac{m}{n}=\log m-\log n

Complete step-by-step solution:
We have given an equation (0.2)x=2{{\left( 0.2 \right)}^{x}}=2.
We have to find the value of xx.
Now, we have to take logarithm on both side to solve further, we get
(0.2)x=2{{\left( 0.2 \right)}^{x}}=2
log⁡(0.2)x=log⁡2\log {{\left( 0.2 \right)}^{x}}=\log 2
Now, we know that log⁡(m)n=nlog⁡m\log {{\left( m \right)}^{n}}=n\log m.
Now, applying to the above equation, we get
xlog⁡(0.2)=log⁡2x\log \left( 0.2 \right)=\log 2
We have given that log⁡102=0.3010{{\log }_{10}}2=0.3010, when we substitute the value we get
xlog⁡(0.2)=0.3010x\log \left( 0.2 \right)=0.3010
xlog⁡(210)=0.3010x\log \left( \dfrac{2}{10} \right)=0.3010
Now, we know that log⁡mn=log⁡m−log⁡n\log \dfrac{m}{n}=\log m-\log n
Now, applying to above equation, we get
x(log⁡2−log⁡10)=0.3010x\left( \log 2-\log 10 \right)=0.3010
Now, we put the value of log⁡2\log 2 and log⁡10\log 10 in the above equation, we get
x(0.3010−1)=0.3010 [As log10=1] x(−0.699)=0.3010 \begin{aligned} & x\left( 0.3010-1 \right)=0.3010\text{ }\left[ \text{As log10=1} \right] \\\ & x\left( -0.699 \right)=0.3010 \\\ \end{aligned}
We have to cross multiply to obtain the value of xx, then value of xx will be

& x=\dfrac{0.3010}{-0.699} \\\ & x=-0.43 \\\ \end{aligned}$$ **So, the value of x is $-0.43$.** **Note:** As given in the question ${{\log }_{10}}2=0.3010$, so we use the base 10 for calculating the value of $\log 10$. There are two logarithms in mathematics, one is a natural logarithm and the other is a common logarithm. The base of the common logarithm is $10$, the base of natural logarithm is $e$ and written as ${{\log }_{e}}x$. Here, $e$ represents a fixed irrational number approximately equal to $2.71828$. When the base of the logarithm is not specified in the question, we generally use a common logarithm.