Question

логарифмы решите 1 и 4

Answer 1

$1)\; \; lg2=a\\\\lg25=lg5^2=2\cdot lg5=2\cdot lg\frac{10}{2}=2\cdot (lg10-lg2)=2\cdot (1-a)\\\\2)\; \; log_{a}b=4\\\\log_{\sqrt{ab}}\Big (\frac{\sqrt{b}}{\sqrt[4]{a}}\Big )+log_{\sqrt{ab}}(a\sqrt{a})=log_{\sqrt{ab}}\; \frac{\sqrt{b}\cdot a\sqrt{a}}{\sqrt[4]{a}}=log_{\sqrt{ab}}\; \Big (b^{\frac{1}{2}}\cdot a^{\frac{3}{2}-\frac{1}{4}}\Big )=\\\\=log_{\sqrt{ab}}\; b^{\frac{1}{2}}+log_{\sqrt{ab}}\; a^{\frac{5}{4}}=\frac{1}{2}\, log_{\sqrt{ab}}\; b+\frac{5}{4}\, log_{\sqrt{ab}}\; a=$

$=\frac{1}{2}\cdot \frac{log_{a}b}{log_{a}\sqrt{ab}}+\frac{5}{4}\cdot \frac{log_{a}a}{log_{a}\sqrt{ab}}=\frac{1}{2}\cdot \frac{4}{log_{a}a^{\frac{1}{2}}+log_{a}b^{\frac{1}{2}}}+ \frac{5}{4}\cdot \frac{1}{log_{a}a^{\frac{1}{2}}+log_{a}b^{\frac{1}{2}}}=\\\\=\frac{2}{\frac{1}{2}+\frac{1}{2}\cdot log_{a}b}+\frac{5}{4}\cdot \frac{1}{\frac{1}{2}+\frac{1}{2}\cdot log_{a}b}=\frac{2}{\frac{1}{2}(1+4)}+\frac{5}{4}\cdot \frac{1}{\frac{1}{2}(1+4)}=\frac{4}{5}+\frac{5}{4}\cdot \frac{2}{5}=$

$=\frac{4}{5}+\frac{1}{2}=\frac{13}{10}=1,3$