Q.Let $b>0$ and $b\ne1$. Express $y=b^x$ in logarithmic form. Also state the domain and range of the logarithmic function.

For a fixed base $0<a\ne1$, the logarithmic function $\log_a(\cdot)$ is defined as the inverse of the exponential $a^x$: $y=a^x\iff\log_ay=x$. Since $a^x$ has domain $R$ and range $(0,\infty)$, $\log_a$ has domain $(0,\infty)$ and range $R$ -- logarithms of non-positive numbers are never defined. Since $a^0=1$ always, $\log_a1=0$ for every base.

The five core laws (all provable directly from the inverse relationship):

Negative-base trick. $\log_{1/a}x=-\log_ax$ (immediate from the change-of-base formula with $\log_a(1/a)=-1$) -- very useful for combining mixed-base logarithmic equations into a single base.

Telescoping chains. A product like $\log_ab\cdot\log_bc\cdot\log_ca$ always collapses to $1$ (change every factor to a common base and watch every intermediate term cancel); more generally $\log_a b_1\cdot\log_{b_1}b_2\cdots\log_{b_{k-1}}b_k=\log_ab_k$.

Named bases. Base $10$ = common logarithm; base $e\approx2.71828$ (Euler's number, the limit of $(1+1/n)^n$ as $n\to\infty$, arising from continuous compounding) = natural logarithm, $\ln x$; base $2$ = binary logarithm, used in computer science.

Solving logarithmic equations. Convert to exponential form to remove the log entirely (e.g. $\log_{5-x}(x^2-6x+65)=2\Rightarrow(5-x)^2=x^2-6x+65$), or combine several logs into one via the product/quotient/power rules first; ALWAYS check the resulting solution against the base condition ($>0$, $\ne1$) and the argument condition ($>0$) before accepting it.