Class 9th math 2.4 solution english PCTB

Question 4: Find the value of \(x\) in the following equations:

\((i)\) \(\quad\) \(log\ 2+log\ x=1 \)

Solution:

\[
\begin{align*}
\log 2+\log x&=1\\
\log x&=1-\log 2\\
\log x&=1-0.3010\\
\log x&=0.699
\end{align*}
\]

Taking antilog on both sides, we have

\[
\begin{align*}
x&=\text{antilog}( 0.699)\\
x&=5
\end{align*}
\]

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