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Logarithms — MATHEMATICS STD 9 — Question

ICSE BoardEnglish MediumSTD 9MATHEMATICSLogarithms5 Marks
Question
Prove that : $2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log 2$
✓

Answer

We need to prove that
$2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}=\log 2$
$ \text { LHS }=2 \log \frac{15}{18}-\log \frac{25}{162}+\log \frac{4}{9}$
$ =\log \left(\frac{15}{18}\right)^2-\log \left(\frac{25}{162}\right)+\log \left(\frac{4}{9}\right) \ldots .\left[\mathrm{n} \log _{\mathrm{a}} \mathrm{m}=\log _{\mathrm{a}} \mathrm{m}^{\mathrm{n}}\right.]$
$ =\log \left[\left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right)\right]-\log \frac{25}{162}+\log \frac{4}{9}$
$ =\log \left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right) \times\left(\frac{4}{9}\right)-\log \left(\frac{25}{162}\right) \ldots . .\left[\log _a m+\log _a n=\log _a(m n)\right]$
$ =\log \frac{\left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right) \times \frac{4}{9}}{\frac{25}{162}} \ldots .\left[\log _a m-\log _a n=\log _a\left(\frac{m}{n}\right)\right]$
$ =\log \left(\frac{15}{18}\right) \times\left(\frac{15}{18}\right) \times \frac{4}{9} \times \frac{162}{25}$
$ =\log \frac{72}{36}$
$ =\log 2$
$ =\text { R.H.S. }$

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