Since \(\displaystyle \log_{10} x = A\) and \(\displaystyle \log_{10} y = B\), it follows that \(\displaystyle 10^{A} = x\) and \(\displaystyle 10^{B} = y\)

\(\displaystyle 100^{2A-B} = \left( 10^{2} \right ) ^{2A-B} =10^{2 (2A-B)} =10^{4A-2B}=\frac{10^{4A}}{10^{2B}} =\frac{\left ( 10^{A}\right )^4}{\left (10^{B}\right )^2}=\frac{x^4}{y^2}\)

First we condense the LHS to \(\displaystyle \ln(3x^2-12)\) using the properties of logarithms.  Then we can eliminate the natural log on the LHS by raising both sides of the equation as exponents with base \(\displaystyle e\). This leaves us with \(\displaystyle 3x^2-12=1\) since \(\displaystyle e^0 = 1\).  Now we do some simple Algebra and obtain our answer.

First take the logarithm of both sides.  This yields \(\displaystyle \log10^{2x}=\log(64)\). Now using the properties of logarithms, we can bring the \(\displaystyle 2x\) down in front of \(\displaystyle \log(10)\). Since \(\displaystyle \log(10)=1\). We are left with \(\displaystyle 2x = \log(64)\).  Dividing each side by \(\displaystyle 2\) we find \(\displaystyle x = \frac{1}{2}\log(64) = \log(\sqrt{64})=\log(8)\).

First, we rewrite the equation in logarithmic form to obtain \(\displaystyle 2x+1=\ln100\). Simplifying, we get

\(\displaystyle 2x = ln100 - 1\)

\(\displaystyle x=\frac{ln100-1}{2}\)

\(\displaystyle x\approx1.803\)

\(\displaystyle \log_464\rightarrow \log_44^3\rightarrow 3\)

Logs must be used to solve this problem. Follow the steps below:

\(\displaystyle 4^x=16^{x-2}\)

\(\displaystyle 4^x=4^{2(x-2)}\)

\(\displaystyle \log_{4}4^x=\log_{4}4^{2(x-2)}\)

\(\displaystyle x=2(x-2)\)

\(\displaystyle x=2x-4\)

\(\displaystyle -x=-4\)

\(\displaystyle x=4\)

First we condense the LHS to \(\displaystyle \log_{2}(p^2-7p)\). Now we can eliminate a log of base \(\displaystyle 2\) by making both sides of the equation exponents of base \(\displaystyle 2\). This leaves us with \(\displaystyle p^2-7p=8\). Moving \(\displaystyle 8\) to the LHS we obtain a simple quadratic which we solve to obtain \(\displaystyle 8\) and \(\displaystyle -1\) for solutions. Finally, we eliminate \(\displaystyle -1\) as logarithms are undefined for negative numbers.

The logarithmic function is undefined when the inputs are negative or 0. Therefore the inputs of the logarithmic function must be positive. This means that the quantity \(\displaystyle 8-4x\) must be positive. After setting up the appropriate inequality, we have,

\(\displaystyle 8-4x>0\)

\(\displaystyle -4x>-8\)

\(\displaystyle x< 2\)

Therefore the domain of the function \(\displaystyle f(x)=\)  \(\displaystyle \log_{4}(8 - 4x)\) is the interval \(\displaystyle (-\infty,2)\).

Here we employ the use of the logarithm base change formula 

\(\displaystyle \log_bx= \frac{\log_cx}{\log_cb}\).

Therefore we get,

\(\displaystyle \log_38=\frac{\log_48}{\log_43}\)

The logarithmic function is undefined when the inputs are negative or 0. Therefore the inputs of the logarithmic function must be positive. This means that the quantity \(\displaystyle 1+\sqrt{-6-3x\) must be positive. However, this quantity is always positive because the value under the radical cannot be negative. So we are only constrained by:

\(\displaystyle \sqrt{-6-3x} \geq 0\)

\(\displaystyle -6-3x \geq0\)

\(\displaystyle -3x\geq6\)

\(\displaystyle x\leq-2\)

Therefore the domain of the function \(\displaystyle f(x)=\log_{10}(1+\sqrt{-6-3x})\) is the interval \(\displaystyle (-\infty,-2]\).