Let $a$ be first term $a=\sqrt[\ln 4]{4}$ which become $a=4^{\frac{1}{\ln 4}}$. Applying $\ln$ on both sides and using power of power rule yields $\ln a=\Big(\frac{1}{\ln 4}\Big)(\ln 4)$ or $\ln a=1$ thus $a=1$.
Using the same analogy, note that all terms are $e$ therefore $\displaystyle \sqrt[\ln 4]{4}\sqrt[\ln 9]{9}\sqrt[\ln 16]{16}\cdot \sqrt[\ln 100]{100}=e \cdot e \cdot e \cdots e=e^9$

Using the formula for the logarithm of a power and a calculator, we have $$\log 6^{3000}=3000\log 6\simeq 2334.45$$ Thus $6^{3000}$ has 2335 digits.

This problem is asking to find an integer $k$ such that $$56 \leq \log 18^{k} < 57$$ Using the power property of logarithm, we can rewrite the inequalities above as $$56\leq k\log 18 < 57$$ Dividing all sides of inequality by log 18 gives $$\frac{56}{\log 18}\leq k < \frac{57}{\log 18}$$ Using a calculator, $$44.61 \leq k < 45.40$$ Thus the only possible choice is $k=45$. Again using a calculator, we see that $$\log 18^{45}=45\log 18 \simeq 56.48$$ Thus $18^{45}$ has 57 digits.

Taking $\log$ on both sides of $5^n > 10^{100}$, we have $$\log 5^n > \log 10^{100}$$ which can be rewritten as $$n \cdot \log 5 >100 \cdot \log 10$$ This implies $$n > \frac{100}{\log 5} \simeq 143.06$$ The smallest integer that is bigger than 143.06 is $n=144$.

\begin{align*} \log_{a\sqrt[5]{a}} a^{\frac{4}{3}} &=\frac{\log a^{\frac{4}{3}}}{\log a\sqrt[5]{a}} &&\text{Change of base}\\ &=\frac{\log a^{\frac{4}{3}}}{\log a^1\cdot a^{\frac{1}{5}}} &&\text{Rational exponents}\\ &= \frac{\frac{4}{3}\log a}{\frac{6}{5}\log a} &&\text{Properties of logarithms}\\ &=\frac{4}{3}\div \frac{6}{5} &&\text{Simplify}\\ &=\frac{20}{18}=\frac{10}{9}\\ \end{align*}

\begin{align*} \log_8 \Big(\frac{64}{\sqrt{x+1}}\Big) &= \log_8 64 -\log_8 (x+1)^{1/2} &&\text{Quotient Rule}\\ &= 2 -\frac{1}{2}\log_8(x+1) &&\text{Power rule} \end{align*}

\begin{align*} \ln \Big(\frac{ab}{\sqrt[3]{c}}\Big) & =\ln (ab) -\ln \sqrt[3]{c}&&\text{Quotient Rule}\\ &=\ln a+\ln b-\ln c^{1/3}&&\text{Product Rule}\\ &=\ln a+\ln b-\frac{1}{3}\ln c &&\text{Power rule} \end{align*}

\begin{align*} \log_2\Big(\frac{z^3}{xy^5}\Big)&=\log_2 z^3-\log_2(xy^5) &&\text{Quotient Rule}\\ &=\log_2 z^3-(\log_2 x +\log_2y^5) &&\text{Product Rule}\\ &=3\log_2 z -\log_2 x-5 \log_2 y &&\text{Power rule} \end{align*}

\begin{align*} \log \sqrt[3]{\frac{(x+y)^2}{10}} &=\log \Big[\frac{(x+y)^2}{10}\Big]^{\frac{1}{3}} && \text{Rational Exponent}\\ &=\frac{1}{3}\log \Big[\frac{(x+y)^2}{10}\Big]&&\text{Power rule}\\ &=\frac{1}{3}[\log(x+y)^2-\log10]&&\text{Quotient Rule}\\ &=\frac{1}{3}[2\log(x+y)-1]&&\text{Power rule}\\ &=\frac{2}{3}\log(x+y)-\frac{1}{3} &&\text{Distributive Property} \end{align*}

\begin{align*} \displaystyle \log_2 560-\log_2 7-\log_2 5 &=\log_2 560 -(\log_2 7 +\log_2 5) &&\text{Factor out (-1)}\\ &=\log_2 560 -\log_2(7 \cdot 5)&&\text{Product Rule}\\ &=\log_2\Big(\frac{560}{7 \cdot 5}\Big)&&\text{Quotient Rule}\\ &=\log_2 16=4 \end{align*}

\begin{align*} \displaystyle 3\ln s +\frac{1}{2}\ln t - 4\ln(t^2+1)&=\ln s^3+\ln t^{1/2}-\ln(t^2+1)^4 &&\text{Product Rule}\\ &=\ln(s^3t^{1/2})-\ln(t^2+1)^4 &&\text{Product Rule}\\ &=\ln\Big(\frac{s^3\sqrt{t}}{(t^2+1)^4}\Big) &&\text{Quotient Rule}\\ \end{align*}

\begin{align*} &\log x+\log(x^2-1)-\log 7-\log(x+1)\\ &=\log x +\log(x^2-1)-(\log 7 +\log(x+1)) &&\text{Factor out (-1)}\\ &= \log(x(x^2-1))-\log(7(x+1)) &&\text{Product Rule}\\ &=\log \frac{x(x^2-1)}{7(x+1)}&&\text{Quotient Rule}\\ &=\log \frac{x(x+1)(x-1)}{7(x+1)}&&\text{Factor difference of squares}\\ &= \log \frac{x(x-1)}{7} &&\text{Simplify} \end{align*}

\begin{align*} &\frac{1}{2}(\log_5 x+\log_5 y)-2\log_5(x+1) \\ &=\frac{1}{2}\log_5 xy-\log_5(x+1)^2 &&\text{Product and Power Rule}\\ &=\log_5(xy)^{1/2}-\log_5(x+1)^2 &&\text{Power Rule}\\ &=\log_5 \frac{(xy)^{1/2}}{(x+1)^2}&&\text{Quotient Rule}\\ &=\log_5 \frac{\sqrt{xy}}{(x+1)^2} \end{align*}

\begin{align*} & \frac{1}{3}[2\ln(x+5)-\ln x -\ln(x^2-4)]\\ &=\frac{1}{3}[\ln(x+5)^2-\ln x -\ln (x^2-4)] &&\text{Power Rule}\\ &=\frac{1}{3}\Big[\ln \frac{(x+5)^2}{x(x^2-4)}\Big] &&\text{Quotient Rule}\\ &=\ln \Big[\frac{(x+5)^2}{x(x^2-4)}\Big]^{1/3} &&\text{Power Rule}\\ &=\ln \sqrt[3]{\frac{(x+5)^2}{x(x^2-4)}} \end{align*}

\begin{align*} \log_4(8x) & =\log_4x+\log_4 2+1\\ \log_4(8x) &= \log_4x+\log_4 2+\log_4 4 && \text{ replace } 1 \text{ with }\log_4 \\ \log_4(8x) & =\log_4(8x) && \text{ by property of logarithms}\\ 8x&= 8x \end{align*} Note that last statement is true for any value of $x$. In our case we limit to $x>0$ which is the domain or our $\log_4 x$ function.

\begin{align*} \Big(\frac{2}{3}\Big)^{k-1}&=\Big(\frac{81}{16}\Big)^{k+1} &&\\ \Big(\frac{2}{3}\Big)^{k-1}&=\Big(\frac{3^4}{2^4}\Big)^{k+1} &&\text{Express } 81 \text{ as } 3^4 \text{ and } 16 \text{ as } 2^4\\ \Big(\frac{2}{3}\Big)^{k-1} &=\Big[\Big(\frac{3}{2}\Big)^4\Big]^{k+1} &&\text{Apply property of exponents } \Big(\frac{a}{b}\Big)^m=\frac{a^m}{b^m}\\ \Big(\frac{2}{3}\Big)^{k-1}&=\Big(\frac{3}{2}\Big)^{4(k+1)} &&\text{Apply power of power property } (a^n)^p=a^{n \cdot p}\\ \Big[\Big(\frac{3}{2}\Big)^{-1}\Big]^{k-1}&=\Big(\frac{3}{2}\Big)^{4(k+1)} &&\text{Apply recipricol property } \Big(\frac{a}{b}\Big)^{-1}=\frac{b}{a}\\ \Big(\frac{3}{2}\Big)^{-k+1} &=\Big(\frac{3}{2}\Big)^{4(k+1)} &&\text{Apply power of power property } (a^n)^p=a^{n \cdot p}\\ -k+1 &=4(k+1) &&\text{Equal powers of exponents}\\ -k+1 &=4k+4 &&\text{Distributy property}\\ 5k &=-3 && \text{Simplify like terms}\\ k &=-\frac{3}{5}\\ \end{align*}

\begin{align*} 4^{2015}-4^{2014}-4^{2013}+4^{2012} & =45(2^x) &&\\ 4^{2012}(4^3-4^2-4^1+1) & =45(2^x)&&\text{Factor out }4^{2012}\\ 4^{2012} \cdot 45 & =45(2^x) &&\text{Simplify}\\ 4^{2012} &=2^x &&\text{Divide both sides by 45}\\ (2^2)^{2012} &=2^x &&\text{Write } 4 \text{ as } 2^2\\ 2^{4024} &=2^x &&\text{Apply power of the power rule } (a^n)^p=a^{n \cdot p}\\ 4024&=2x &&\text{Equal powers of exponents}\\ x&=2012 &&\text{Divide both sides by 2}\\ \end{align*}

\begin{align*} & (\ln x)^2-\ln x^4-12=0 && \text{Original Equation}\\ & (\ln x)^2-4\ln x-12=0 && \text{Apply power rule of logarithm } p\log_b M=\log_b M^p\\ & (\ln x-6)(\ln x +2)=0 &&\text{Factoring }\\ &\ln x-6=0 \text{ or } \ln x+2=0 &&\text{By zero product property}\\ &\ln x=6 \text{ or } \ln x=-2 &&\text{Simplify}\\ & x=e^6 \text{ or } x=e^{-2}=\frac{1}{e^2} && \text{Definition of logarithm} \end{align*}

\begin{align*} & \displaystyle \frac{\log 3}{\log 2}\cdot \frac{\log 4}{\log 3}\cdot \frac{\log 5}{\log 4}\cdots \frac{\log(n+1)}{\log n}=3\\ &\text{Note that we can cross simplify most of the terms, which yields }\\ &\displaystyle \frac{\log(n+1)}{\log 2}=3\\ &\displaystyle \log(n+1)=3\cdot \log 2\\ &\displaystyle \log(n+1)=\log 2^3\\ &\displaystyle n+1=8\\ &n=7\\ \end{align*}