Question

If a and b are positive real numbers such that 2 log (3a - 2b) = log a + log b, then the value of $\frac{a}{b}$=

• A. $1$
• B. $\frac{2}{3}$
• C. $\frac{4}{9}$
• D. $2$

Answer 1

The correct option is A $1$

Given that,
$2\log \left(3a-2b\right)=\log a+\log b\mathrm{we}\mathrm{know}\mathrm{that}log\mathrm{can}\mathrm{only}\mathrm{takes}+\mathrm{ve}\mathrm{values}\mathrm{in}\mathrm{its}\mathrm{domain}3a-2b>0\Rightarrow 3a>2b\Rightarrow \frac{a}{b}>\frac{2}{3}(b>0)\Rightarrow \log {\left(3a-2b\right)}^2=\log \left(ab\right)\Rightarrow 4b^2+9a^2-12ab=ab\Rightarrow 4b^2+9a^2-13ab=0\Rightarrow 9{\left(\frac{a}{b}\right)}^2-13\frac{a}{b}+4=0\Rightarrow \left(9\frac{a}{b}-4\right]\left(\frac{a}{b}-1\right]=0\Rightarrow \frac{a}{b}=1or\frac{4}{9}\frac{a}{b}=\frac{4}{9}isnotpossible$