Question

If $(2.3{)}^x$$=(0.23{)}^y=1000$ then find the value of $\frac{1}{x}-\frac{1}{y}$.

Answer 1

Step $1$: Simplify the equation.

$(2.3{)}^x$$=(0.23{)}^y=1000$

Applying ${\log }_{10}$ on the equation

${\log }_{10}{\left(2.3\right)}^x$ $={\log }_{10}(0.23{)}^y={\log }_{10}\left(1000\right)$ $\left[\log a^x=x\log a\right]$

$x{\log }_{10}\left(2.3\right)$$=y{\log }_{10}(0.23)=3$

Step $2$: Solve for $x$.

Taking $x{\log }_{10}\left(2.3\right)$$=3$

$\Rightarrow {\log }_{10}\left(2.3\right)=\frac{3}{x}$……………….$\left(1\right)$

Step $3$: Solve for $y$ :

Taking $y{\log }_{10}\left(0.23\right)$$=3$

$\Rightarrow {\log }_{10}\left(0.23\right)=\frac{3}{y}$……………..$\left(2\right)$

Step $4$ : Find the value of $\frac{1}{x}-\frac{1}{y}$:

From, equation $\left(1\right)$ and equation $\left(2\right)$, we can write that

$$\begin{gathered} \frac{3}{x}-\frac{3}{y}={\log }_{10}\left(2.3\right)-{\log }_{10}\left(0.23\right);\left[{\log }_{10}\left(a\right)-{\log }_{10}\left(b\right)={\log }_{10}\left(\frac{a}{b}\right)\right] \\ 3\left(\frac{1}{x}-\frac{1}{y}\right)={\log }_{10}\left(\frac{2.3}{0.23}\right) \\ 3\left(\frac{1}{x}-\frac{1}{y}\right)={\log }_{10}\left(10\right);\left[{\log }_{10}\left(10\right)=1\right] \\ 3\left(\frac{1}{x}-\frac{1}{y}\right)=1 \\ \frac{1}{x}-\frac{1}{y}=\frac{1}{3} \end{gathered}$$

Hence, the value of $\frac{1}{x}-\frac{1}{y}$ is $\frac{1}{3}$.