Question
Evaluate the following without using tables :$\log 4+\frac{1}{3} \log 125-\frac{1}{5} \log 32$
✓
Answer
Consider the given expression
$\log 4+\frac{1}{3} \log 125-\frac{1}{5} \log 32$
$ =\log 4+\log (125)^{\frac{1}{3}}-\log (32)^{\frac{1}{5}} \ldots\left[n \log _a m=\log _a m^n\right]$
$ =\log 4+\log \left(5^3\right)^{\frac{1}{3}}-\log \left(2^5\right)^{\frac{1}{5}}$
$ =\log 4+\log 5-\log 2$
$ =\log 4 \times 5-\log 2 \ldots .\left[\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \mathrm{mn}\right]$
$ =\log \left(\frac{20}{2}\right) \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log 10$
$ =1$
$\log 4+\frac{1}{3} \log 125-\frac{1}{5} \log 32$
$ =\log 4+\log (125)^{\frac{1}{3}}-\log (32)^{\frac{1}{5}} \ldots\left[n \log _a m=\log _a m^n\right]$
$ =\log 4+\log \left(5^3\right)^{\frac{1}{3}}-\log \left(2^5\right)^{\frac{1}{5}}$
$ =\log 4+\log 5-\log 2$
$ =\log 4 \times 5-\log 2 \ldots .\left[\log _{\mathrm{a}} \mathrm{m}+\log _{\mathrm{a}} \mathrm{n}=\log _{\mathrm{a}} \mathrm{mn}\right]$
$ =\log \left(\frac{20}{2}\right) \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log 10$
$ =1$
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