Question
Find $x$, if $: x-\log 48+3 \log 2=\frac{1}{3} \log 125-\log 3$.
✓
Answer
Consider the given equation
$x-\log 48+3 \log 2=\frac{1}{3} \log 125-\log 3$
$ \Rightarrow x=\frac{1}{3} \log 125-\log 3+\log 48-3 \log 2$
$ \Rightarrow x=$
$ \log (125)^{\frac{1}{3}}-\log 3+\log 48-\log 2^3 \ldots .\left[n \log _a m=\log _a m^n\right]$
$ \Rightarrow x=\log (5 \times 5 \times 5)^{\frac{1}{3}}-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log \left(5^3\right)^{\frac{1}{3}}-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log 5-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log 5+\log 48-\log 3-\log 8$
$ \Rightarrow x=(\log 5+\log 48)-(\log 3+\log 8)$
$ \Rightarrow \mathrm{x}=(\log 5 \times 48)-(\log 3 \times 8) \ldots .\left[\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \mathrm{mn}\right]$
$ \Rightarrow \mathrm{x}=\log \frac{5 \times 48}{3 \times 8} \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ \Rightarrow \mathrm{x}=\log \frac{5 \times 6 \times 8}{3 \times 8}$
$ \Rightarrow \mathrm{x}=\log 10$
$ \Rightarrow \mathrm{x}=1$
$x-\log 48+3 \log 2=\frac{1}{3} \log 125-\log 3$
$ \Rightarrow x=\frac{1}{3} \log 125-\log 3+\log 48-3 \log 2$
$ \Rightarrow x=$
$ \log (125)^{\frac{1}{3}}-\log 3+\log 48-\log 2^3 \ldots .\left[n \log _a m=\log _a m^n\right]$
$ \Rightarrow x=\log (5 \times 5 \times 5)^{\frac{1}{3}}-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log \left(5^3\right)^{\frac{1}{3}}-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log 5-\log 3+\log 48-\log 8$
$ \Rightarrow x=\log 5+\log 48-\log 3-\log 8$
$ \Rightarrow x=(\log 5+\log 48)-(\log 3+\log 8)$
$ \Rightarrow \mathrm{x}=(\log 5 \times 48)-(\log 3 \times 8) \ldots .\left[\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \mathrm{mn}\right]$
$ \Rightarrow \mathrm{x}=\log \frac{5 \times 48}{3 \times 8} \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ \Rightarrow \mathrm{x}=\log \frac{5 \times 6 \times 8}{3 \times 8}$
$ \Rightarrow \mathrm{x}=\log 10$
$ \Rightarrow \mathrm{x}=1$
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