\(\displaystyle 10^{t} + 1 = e\)

\(\displaystyle 10^{t} = e - 1\)

Take the common logarithm of both sides and solve for \(\displaystyle t\):

\(\displaystyle \log 10^{t} = \log \left (e - 1 \right )\)

\(\displaystyle t = \log (e - 1) \approx \log (2.718-1) \approx \log 1.718 \approx 0.24\)

Take the natural logarithm of both sides and solve for \(\displaystyle x\):

\(\displaystyle e^{x} = \pi\)

\(\displaystyle \ln e^{x} = \ln \pi\)

\(\displaystyle x = \ln \pi \approx \ln 3.1416 \approx 1.14\)

Take the common logarithm of both sides, then solve the resulting linear equation.

\(\displaystyle 2^{x+4} = 5^{x}\)

\(\displaystyle \log 2^{x+4} = \log 5^{x}\)

\(\displaystyle (x + 4) \log 2 = x \log 5\)

\(\displaystyle x\log 2 + 4\log 2 = x \log 5\)

\(\displaystyle 4\log 2 = x \log 5 - x\log 2\)

\(\displaystyle 4\log 2 = x\left ( \log 5 - \log 2 \right )\)

\(\displaystyle x = \frac{4\log 2 }{ \log 5 - \log 2 }\)

\(\displaystyle x \approx \frac{4 \cdot 0.3010}{0.6990-0.3010} \approx 3.03\)

Take the common logarithm of both sides, then solve the resulting linear equation.

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

\(\displaystyle \log 2^{x+10}= \log 10^{x+2}\)

\(\displaystyle \left (x+10 \right )\log 2 = x+2\)

\(\displaystyle x\log 2 +10 \log 2 = x+2\)

\(\displaystyle x\log 2 +10 \log 2 - 2 - x \log 2= x+2 - 2 - x \log 2\)

\(\displaystyle -2 +10 \log 2 = x - x \log 2\)

\(\displaystyle x \left (1 - \log 2 \right )= -2 +10 \log 2\)

\(\displaystyle x =\frac{ -2 +10 \log 2}{1 - \log 2 }\)

\(\displaystyle x \approx \frac{ -2 +10 \cdot 0.3010}{1 - 0.3010 } \approx 1.45\)

Take the common logarithm of both sides:

\(\displaystyle 100 ^{x}= 66\)

\(\displaystyle \log 100 ^{x}= \log 66\)

\(\displaystyle x \log 100 = \log 66\)

\(\displaystyle x \cdot 2 = \log 66\)

\(\displaystyle x = \frac{ \log 66}{2}\)

The base of the common logarithm is 10, so 

\(\displaystyle 3 = \log 10^{3} = \log 1,000\)

The sum of three logarithms is the logarithm of the product of the three powers, so:

\(\displaystyle 3 + \log 4 + \log 6\)

\(\displaystyle =\log 1,000 + \log 4 + \log 6\)

\(\displaystyle = \log (1,000 \times 4 \times 6) = \log 24,000\)

Therefore, \(\displaystyle N = 24,000\)

\(\displaystyle 5^{N} = 10^{4-N}\)

\(\displaystyle \log 5^{N} = \log 10^{4-N}\)

\(\displaystyle N \log 5 = 4 - N\)

\(\displaystyle N \log 5 + N = 4\)

\(\displaystyle N \left (1+ \log 5 \right ) = 4\)

\(\displaystyle N = \frac{4}{1+ \log 5 } \approx \frac{4}{1+ 0.6990 } \approx 2.35\)

The base of the common logarithm is 10, so 

\(\displaystyle 2 = \log 10^{2} = \log 1 00\)

The sum of logarithms is the logarithm of the product of the three powers, and the difference of logarithms is the logarithm of the quotient of their powers. Therefore, 

\(\displaystyle \log 20 - 2 + \log 6\)

\(\displaystyle = \log 20 - \log 100 + \log 6\)

\(\displaystyle = \log \frac{20 \times 6 }{100} = \log \frac{120 }{100} =\log \frac{6}{5}\)

\(\displaystyle N = \frac{6}{5} = 1 \frac{1}{5}\)

Using the Product of Powers Rule, then the Power of a Power Rule, rewrite the first two terms strategically:

\(\displaystyle 2^{2( x+2)} - 2 ^{(x+2)+ 2 } + 3 = 0\)

\(\displaystyle (2^{ x+2} ) ^{2 }-2 ^ {2} \cdot 2 ^{x+2 } + 3 = 0\)

\(\displaystyle (2^{ x+2} ) ^{2 }-4 ( 2 ^{x+2 }) + 3 = 0\)

Substitute \(\displaystyle t\) for \(\displaystyle 2^{x+2 }\); the equation becomes

\(\displaystyle t ^{2 }-4t + 3 = 0\)

Factor this as 

\(\displaystyle (t+ \square)(t+ \square) = 0\) 

by finding two integers whose product is 3 and whose sum is \(\displaystyle -4\). Through some trial and error we find \(\displaystyle -1, -3\), so we can write

\(\displaystyle (t-1)(t-3) = 0\)

By the Zero Product Rule, one of these two factors must be equal to 0.

If \(\displaystyle t- 1 = 0\), then \(\displaystyle t = 1\).

Substituting \(\displaystyle 2^{x+2 }\) back for \(\displaystyle t\):

\(\displaystyle 2^{x+2 } = 1\)

\(\displaystyle 2^{x+2 } = 2^{0}\)

\(\displaystyle x+2 = 0\)

\(\displaystyle x = -2\).

If \(\displaystyle t- 3 = 0\), then \(\displaystyle t = 3\).

Substituting \(\displaystyle 2^{x+2 }\) back for \(\displaystyle t\):

\(\displaystyle 2^{x+2 } = 3\)

\(\displaystyle \ln 2^{x+2 } = \ln 3\)

\(\displaystyle (x+2 )\ln 2 = \ln 3\)

\(\displaystyle x \ln 2 +2 \ln 2 = \ln 3\)

\(\displaystyle x \ln 2 +2 \ln 2- 2 \ln 2 = \ln 3 - 2 \ln 2\)

\(\displaystyle x \ln 2 = \ln 3 - 2 \ln 2\)

\(\displaystyle \frac{x \ln 2 }{\ln 2}=\frac{ \ln 3 - 2 \ln 2}{\ln 2}\)

\(\displaystyle x \approx \frac{1.0986 - 2 (0.6931 )}{ 0.6931 }\)

\(\displaystyle x \approx -0.42\)

Both can be confirmed to be solutions by substitution.

Rewrite by taking advantage of the Product of Powers Property and the Power of a Power Property:

\(\displaystyle 6^{2x+1} + 6 = 6 ^{x+2} + 6 ^{x}\)

\(\displaystyle 6^{1 } \cdot 6^{2x} + 6 = 6 ^{2} \cdot 6 ^{x } + 6 ^{x}\)

\(\displaystyle 6 \cdot 6^{2x} + 6 =36 \cdot 6 ^{x } + 6 ^{x}\)

\(\displaystyle 6 \cdot (6^{x } )^{2 }+ 6 = 37 \cdot( 6 ^{x })\)

Substitute \(\displaystyle t\) for \(\displaystyle 6^{x}\); the resulting equation is the quadratic equation

\(\displaystyle 6 t ^{2 }+ 6 = 37t\)

which can be written in standard form by subtracting \(\displaystyle 37t\) from both sides:

\(\displaystyle 6 t ^{2 }+ 6 - 37t = 37t - 37t\)

\(\displaystyle 6 t ^{2 }- 37t + 6 = 0\)

The trinomial can be factored by the \(\displaystyle ac\) method, Look for two integers with sum \(\displaystyle -37\) and product \(\displaystyle 6 \cdot 6 = 36\); by trial and error, we find they are \(\displaystyle -1, -36\), so the equation can be rewritten and solved by grouping:

\(\displaystyle 6 t ^{2 }- 36t - t + 6 = 0\)

\(\displaystyle (6 t ^{2 }- 36t) - (t - 6) = 0\)

\(\displaystyle 6t (t - 6) - 1 (t - 6) = 0\)

\(\displaystyle (6t - 1 )(t - 6) = 0\)

By the Zero Product Property, one of these factors must be equal to 0. 

Either

\(\displaystyle 6t - 1 = 0\)

\(\displaystyle 6t - 1 + 1 = 0 + 1\)

\(\displaystyle 6t = 1\)

\(\displaystyle \frac{6t}{6} = \frac{1}{6}\)

\(\displaystyle t = \frac{1}{6}\)

Substituting \(\displaystyle 6^{x}\) back for \(\displaystyle t\):

\(\displaystyle 6^{x} = \frac{1}{6}\)

\(\displaystyle 6^{x} = 6^{-1 }\)

\(\displaystyle x = -1\)

Or:

\(\displaystyle t- 6 = 0\)

\(\displaystyle t- 6 + 6 = 0 + 6\)

\(\displaystyle t = 6\)

Substituting \(\displaystyle 6^{x}\) back for \(\displaystyle t\):

\(\displaystyle 6^{x} = 6\)

\(\displaystyle 6^{x} = 6^{1}\)

\(\displaystyle x = 1\)

The solution set, as can be confirmed by substituting in the equation, is \(\displaystyle \left \{ -1, 1 \right \}\).