The first thing we notice about this problem is that \(\displaystyle x\) is an exponent. This should be an immediate reminder: use logs!

The question is, which base should we choose for the log? We should use the natural log (log base e) because the right-hand side of the equation already has e as a base of an exponent. As you will see, things cancel out more nicely this way.

\(\displaystyle 5^{x}=2e^5\)

Take the natural log of both sides:

\(\displaystyle \textup{ln}(5^x) = \textup{ln}(2e^5)\)

Rewrite the right-hand side of the equation using the product rule for logs:

\(\displaystyle \textup{ln}5^x = \textup{ln}2+\textup{ln}e^5\)

Now rewrite the whole equation after bringing down those exponents.

\(\displaystyle x\textup{ln}5 = \textup{ln}2+5\textup{ln}e\)

\(\displaystyle \textup{ln}e\) is the same thing as \(\displaystyle \textup{log}_{e}e\), which equals 1.

\(\displaystyle x\textup{ln}5 = \textup{ln}2+5\)

Now we just divide by \(\displaystyle \textup{ln}5\) on both sides to isolate \(\displaystyle x\).

\(\displaystyle x = \frac{\textup{ln}2+5}{\textup{ln}5}\)

The original equation is:

\(\displaystyle 6e^{-x}+1=3\)

Subtract \(\displaystyle 1\) from both sides:

\(\displaystyle 6e^{-x}=2\)

Divde both sides by \(\displaystyle 6\):

\(\displaystyle e^{-x}=\frac{2}{6}=\frac{1}{3}\)

Take the natural logarithm of both sides:

\(\displaystyle -x=\ln \frac{1}{3}\)

Divde both sides by \(\displaystyle -1\) and use a calculator to get:

\(\displaystyle x=-1.099\)

Give both sides the same base, using e:

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

Because e and ln cancel each other out, \(\displaystyle x+3=e^3\).

Solve for x and round to the nearest tenth:

\(\displaystyle x=e^3-3\)

\(\displaystyle x\approx17.1\)

To solve this exponential function, you must first "undo" the e by taking the natural log of both sides. Thus,

\(\displaystyle \ln(e^{x+2})=\ln3\)

\(\displaystyle x+2=\ln3\)

Then, to isolate x, you must subtract two from both sides. Thus,

\(\displaystyle x=\ln3-2\)

To solve, you must isolate x. The first two steps are to divide both sides by 2 and then take the natural log of both sides.

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

\(\displaystyle \ln{e^{4x}}=\ln2\)

\(\displaystyle 4x=\ln2\)

Finally, you must divide both sides by 4.

\(\displaystyle \frac{4x}{4}=\frac{\ln2}{4}\)

\(\displaystyle x=\frac{\ln2}{4}\)

Case 1,

 \(\displaystyle a= b\)

 \(\displaystyle a^{10x-1}=b^{10x-1}\)

\(\displaystyle a^{10x-1}=a^{10x-1}\)

\(\displaystyle 10x-1 = 10x-1\)

\(\displaystyle x=x\)

This is true for all real values of \(\displaystyle x\), therefore, 

\(\displaystyle x=\left \{x| -\infty < x < \infty \right \}\)

Case 2

 \(\displaystyle a \neq b\)

\(\displaystyle a^{10x-1}=b^{10x-1}\)

Take the natural logarithm of both sides (note that we could use any logarithm but it's convenient to just choose the natural logarithm). 

\(\displaystyle \ln\left(a^{10x-1} \right ) = \ln\left[b^{10x-1}\right]\)

Use the rule for pulling out exponents \(\displaystyle \ln(u^c)=c\ln(u)\). 

\(\displaystyle (10x-1)\ln(a) = (10x-1)\ln(b)\)

Note that if you were to divide out by \(\displaystyle 10x-1\), you would obtain \(\displaystyle \ln(a)=\ln(b)\) which would imply \(\displaystyle a= b\) which is not true for this case. We are solving for \(\displaystyle a\neq b\). 

Expand with the distributive property, 

\(\displaystyle 10x\ln(a)-\ln(a)=10x\ln(b)-\ln(b)\)

Collect and isolate terms with \(\displaystyle x\) onto one side of the equation, 

 \(\displaystyle 10x\ln(a)-10x\ln(b)=\ln(a)-\ln(b)\)

Factor out \(\displaystyle 10x\) (you could just factor out \(\displaystyle x\) and leave the \(\displaystyle 10\) in front of the logarithms but it's easier to see the solution writing it this way).

\(\displaystyle 10x[\ln(a)-\ln(b)]=\ln(a)-\ln(b)\)

\(\displaystyle 10x = \frac{\ln(a)-\ln(b)}{\ln(a)-\ln(b)}\)

\(\displaystyle 10x =1\)

\(\displaystyle x = \frac{1}{10}\)

What's remarkable about this solution is that it was obtained without specifying any value for \(\displaystyle a\) or \(\displaystyle b\). The solution is true so as long as \(\displaystyle a \neq b\). 

Solve for \(\displaystyle y\), 

\(\displaystyle 7(5^{y+1})=7^y\)

Note the rules for logarithms, 

\(\displaystyle \ln(ab) = \ln(a)+\ln(b)\)                                             (1)

\(\displaystyle \ln\left(\frac{a}{b} \right )= \ln(a)-\ln(b)\)                                           (2)

\(\displaystyle \ln(a^n)=n\ln(a)\)                                                        (3)

There is more than one way to get started, but notice that if we divide out by \(\displaystyle 7\) we can proceed as follows, 

\(\displaystyle 5^{y+1}=7^{y-1}\)

Take the natural log of both sides, 

\(\displaystyle \ln(5^{y+1})=\ln(7^{y-1})\)

Use rule (3) to pull out the exponents, 

\(\displaystyle (y+1)\ln(5)=(y-1)\ln(7)\)

Expand with the distributive property, 

\(\displaystyle y\ln(5)+\ln(5)= y\ln(7)-\ln(7)\)

Collect terms with \(\displaystyle y\) on onside and then factor out \(\displaystyle y\). 

\(\displaystyle y[\ln(5)-ln(7)]=-ln(5)-ln(7)\)

Divide out to solve for \(\displaystyle y\), 

\(\displaystyle y = \frac{-\ln(5)-\ln(7)}{\ln(5)-\ln(7)}\)                                                      (4)

The numerator can be simplified using rules (2) and (3). Use rule (3) on the \(\displaystyle -\ln(5)\) term in the numerator as follows, 

\(\displaystyle -\ln(5)= (-1)\ln(5)=\ln(5^{-1})=\ln\left(\frac{1}{5} \right )\)

So now the numerator can be written, 

\(\displaystyle \ln\left(\frac{1}{5} \right ) -\ln(7) = \ln\left(\frac{1/5}{7} \right ) = \ln\left(\frac{1}{35} \right )=-\ln(35)\)

Bringing back the denominator in equation (4) gives, 

\(\displaystyle y=-\frac{ \ln(35) }{\ln(5)-\ln(7)}\) 

\(\displaystyle y \approx 10.57\)

To solve \(\displaystyle 2^{x+4}=512\), use the one-to-one property of exponential equations.

\(\displaystyle 2^{x+4}=2^9\)

\(\displaystyle x+4=9\)

\(\displaystyle \boldsymbol{x=5}\)

For this problem, we can use the fact that the exponent and logarithm are inverses of each other. So if we raise both sides up as exponents:

\(\displaystyle 3^{\log_3(a)} = 3^5\)

The left side of the equation cancels and we get that

\(\displaystyle a = 3^5 = 243\)

To more easily solve for x, we need to try to express the right hand side of the equation in terms of a number of base 4. 256 is simply \(\displaystyle 4^{4}\).

Rewriting the equation with this discovery yields

\(\displaystyle 4^{3x+1}=4^{4}\)

Since both bases are the same, we can set the exponents equal to each other and solve for x.

\(\displaystyle 3x+1=4\)

\(\displaystyle 3x=3\)

\(\displaystyle x=1\)