The derifintion of logarithm is:

\(\displaystyle log_{a}y=x \Leftrightarrow a^{x}=y.\)

In this problem,

\(\displaystyle a= 10, y=1000\) 

\(\displaystyle 10^{3}=1000\)

Therefore, \(\displaystyle x=3\)

Using the rules of logarithms, 

\(\displaystyle \log a - \log b = \log \bigg(\frac{a}{b}\bigg)\)

Hence,

 \(\displaystyle \log x^2 - \log 2x = log \bigg(\frac{x^2}{2x}\bigg) = log \bigg(\frac{x}{2}\bigg) = 2\)

So exponentiate both sides with a base 10:

\(\displaystyle 10^{log_{10} \frac{x}{2}} = 10^{2}\)

The exponent and the logarithm cancel out, leaving:  

\(\displaystyle \frac{x}{2}=10^2\)

\(\displaystyle x=100\cdot 2=200\)

This answer does not match any of the answer choices, therefore the answer is 'None of the other choices'.

In order to solve this equation, we must apply several properties of logarithms. First we notice the term on the left side of the equation, which we can rewrite using the following property:

\(\displaystyle a\log_b (x)=\log_b (x^a)\)

Where a is the coefficient of the logarithm and b is some arbitrary base. Next we look at the right side of the equation, which we can rewrite using the following property for the addition of logarithms:

\(\displaystyle \log _ba+\log _bc=\log_b (a\cdot c)\)

Using both of these properties, we can rewrite the logarithmic equation as follows:

\(\displaystyle 2\log _3(x)=\log _3(4)+\log _3(x-1)\)

\(\displaystyle \log _3(x^2)=\log _3(4(x-1))\)

We have the same value for the base of the logarithm on each side, so the equation then simplifies to the following:

\(\displaystyle x^2=4x-4\rightarrow x^2-4x+4=0\)

Which we can then factor to solve for \(\displaystyle x\):

\(\displaystyle (x-2)(x-2)=0\rightarrow x=2\)

We solve the equation as follows:

\(\displaystyle \ln \frac{cx}{ab+c} = b+c\)

Exponentiate both sides.

\(\displaystyle \frac{cx}{ab+c}= e^{b+c}\) 

Apply the power rule on the right hand side.

\(\displaystyle \frac{cx}{ab+c} = e^be^c\) 

Multiply by \(\displaystyle ab+c\).

\(\displaystyle cx = e^be^c(ab+c)\) 

Divide by \(\displaystyle c\).

\(\displaystyle x =\frac{e^be^c(ab+c)}{c}\)

First, simplify the logarithmic expressions on the left side of the equation:

\(\displaystyle 2log(x)\) can be re-written as \(\displaystyle log(x^2)\).

Now we have:

\(\displaystyle log(27)-log(x^2)=log(x)\).

The left can be consolidated into one log expression using the subtraction rule:

\(\displaystyle log\left(\frac{27}{x^2}\right)=log(x)\).

We now have log on both sides, so we can be confident that whatever is inside these functions is equal:

\(\displaystyle \frac{27}{x^2}=x\) to continue solving, multiply by \(\displaystyle x^2\) on both sides:

\(\displaystyle 27 = x^3\) take the cube root:

\(\displaystyle 3 = x\)

First bring the inside exponent in front of the natural log.

\(\displaystyle (\ln (x^2))^2=\ln (x^4)+3\rightarrow(2\ln (x))^2=4\ln (x)+3\).

Next simplify the first term and bring all the terms on one side of the equation. 

\(\displaystyle 4(\ln (x))^2-4\ln (x)-3=0\).

Next, let set 

\(\displaystyle \ln (x)=A\), so \(\displaystyle 4A^2-4A-3=0\).

Now use the quadratic formula to solve for \(\displaystyle A\). 

\(\displaystyle \\ \frac{-b\pm \sqrt{b^2-4(a)(c)}}{2a}\\ \\=\frac{4\pm\sqrt{16-4(4)(-3)}}{2(4)}\\ \\=\frac{4\pm \sqrt{16+48}}{8}\\ \\=4\pm \frac{\sqrt{64}}{8}\\ \\=\frac{4\pm 8}{8}\\ \\=\frac{3}{2}; -\frac{1}{2}\) 

and thus, \(\displaystyle A=-\frac{1}{2}\) and \(\displaystyle A=\frac{3}{2}\).

Now substitute \(\displaystyle A\) with \(\displaystyle \ln (x)\).

So, \(\displaystyle \ln (x)=-\frac{1}{2}\rightarrow \O\) since \(\displaystyle x\nleqslant0\) and \(\displaystyle \ln (x)=\frac{3}{2}\rightarrow x=e^{3/2}\).

Thus, \(\displaystyle x=e^{3/2}\). 

\(\displaystyle \ln\sqrt{x+2}=2\)

Exponentiate each side to cancel the natural log:

\(\displaystyle e^{\ln\sqrt{x+2}}=e^2\)

\(\displaystyle \sqrt{x+2}=e^2\)

Square both sides:

\(\displaystyle (\sqrt{x+2})^2=(e^2)^2\)

\(\displaystyle x+2=e^4\)

Isolate x:

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

The base of a logarithm is 10 by default:

\(\displaystyle \log _{10 } (2 x + 1 ) = 3\) convert to exponent to isolate x

\(\displaystyle 10^3 = 2x + 1\)

\(\displaystyle 1,000 = 2x + 1\) subtract 1 from both sides

\(\displaystyle 999 = 2x\) divide both sides by 2

\(\displaystyle 499.5 = x\)

First, condense the left side into one logarithm:

\(\displaystyle \log _3 \frac{x}{7 } = 2\) convert to an exponent

\(\displaystyle 3^2 = \frac{x}{7 }\)

\(\displaystyle 9 = \frac{x}{7}\) multiply both sides by 7

\(\displaystyle 63 = x\)

First, consolidate the left side into one logarithm:

\(\displaystyle \log _4 (3x^2 - 7x ) = 3\) convert to an exponent

\(\displaystyle 3x^2 - 7x = 64\) subtract 64 from both sides

\(\displaystyle 3x^2 - 7 x - 64 = 0\) now we can solve using the quadratic formula:

\(\displaystyle x = \frac{7 \pm \sqrt{49 - 4(3)(-64)}}{6 } \approx 5.93, -3.60\)