Question

answer the 3, 4, 5 fast​

Answer 1

3.  We see that,

$\longrightarrow\sf{\log_{x^2}\sqrt{x^2}=\log_{x^2}\left[\left(x^2\right)^{\frac{1}{2}}\right]}$

Since $\sf{\log_b(a^m)=m\,\log_ba,}$

$\longrightarrow\sf{\log_{x^2}\sqrt{x^2}=\dfrac{1}{2}\,\log_{x^2}\left(x^2\right)}$

Since $\sf{\log_aa=1,}$

$\longrightarrow\sf{\log_{x^2}\sqrt{x^2}=\dfrac{1}{2}\times1}$

$\longrightarrow\sf{\underline{\underline{\log_{x^2}\sqrt{x^2}=\dfrac{1}{2}}}}$

4.  Given,

$\longrightarrow\sf{\log_2(5x+3)=\log_2(2x+1)}$

Since $\sf{\log_ba=\dfrac{\log a}{\log b},}$

$\longrightarrow\sf{\dfrac{\log(5x+3)}{\log2}=\dfrac{\log(2x+1)}{\log 2}}$

$\longrightarrow\sf{\log(5x+3)=\log(2x+1)}$

$\longrightarrow\sf{5x+3=2x+1}$

$\longrightarrow\sf{3x=-2}$

$\longrightarrow\sf{x=-\dfrac{2}{3}}$

But, since $\sf{\log x}$ is possible only for $\sf{x\ \textgreater\ 0,}$

$\longrightarrow\sf{5x+3\ \textgreater\ 0\quad AND\quad 2x+1\ \textgreater\ 0}$

$\longrightarrow\sf{x\ \textgreater-\dfrac{3}{5}\quad AND\quad x\ \textgreater-\dfrac{1}{2}}$

Since $\sf{-\dfrac{1}{2}\ \textgreater-\dfrac{3}{5},}$

$\longrightarrow\sf{x\ \textgreater-\dfrac{1}{2}}$

This implies $\sf{x}$ can't be $\sf{-\dfrac{2}{3},}$ because $\sf{-\dfrac{2}{3}\ \textless-\dfrac{1}{2}.}$

Hence there are no possible values for $\sf{x}$ and thus the equation $\sf{\log_2(5x+3)=\log_2(2x+1)}$ has no solution.

5.  Here,

$\longrightarrow\sf{2\ \textless\ 3}$

$\longrightarrow\sf{\dfrac{2\times3}{3}\ \textless\ \dfrac{3\times3}{3}}$

$\longrightarrow\sf{\underline{\underline{\dfrac{6}{3}\ \textless\ \dfrac{7}{3}\ \textless\ \dfrac{8}{3}\ \textless\ \dfrac{9}{3}}}}$

So $\sf{\dfrac{7}{3}}$ and $\sf{\dfrac{8}{3}}$ are two rational numbers between 2 and 3.