To solve $\displaystyle e^l^n^(^2^x^+^3^)$, it is necessary to know the property of $\displaystyle e$.  

$\displaystyle e^l^n^(^x^)=x$

Since $\displaystyle e$ and the $\displaystyle ln$ terms cancel due to inverse operations, the answer is what's left of the $\displaystyle ln$ term.

The answer is:  $\displaystyle 2x+3$

$\displaystyle log{_{a}}^{b}=x$, $\displaystyle b>0$

$\displaystyle \Rightarrow a^{x}=b$

So, $\displaystyle {{log_{4}}^{\frac{1}{2}}}=x\Rightarrow 4^{x}=\frac{1}{2}$

This is a general formula that you should memorize. The inverse of $\displaystyle Log_{b}x=y$ is $\displaystyle b^{y}=x$. You can use this formula to change an equation from a log function to an exponential function.

Rewrite the following expression as an exponential expression:

$\displaystyle log_3b=14,000$

Recall the following property of logs and exponents:

$\displaystyle log_b(a)=c$ 

Can be rewritten in the following form:

$\displaystyle b^c=a$

So, taking the log we are given;

$\displaystyle log_3b=14,000$

We can rewrite it in the form:

$\displaystyle 3^{14,000}=b$

So b must be a really huge number!

Convert the following logarithmic equation to an exponential equation:

$\displaystyle log_{13}b=147$

Recall the following:

This

$\displaystyle log_b(a)=c$

Can be rewritten as

$\displaystyle b^c=a$

So, our given logarithm

$\displaystyle log_{13}b=147$

Can be rewritten as

$\displaystyle 13^{147}=b$

Fortunately we don't need to expand, because this woud be a very large number!

Convert the following logarithmic equation to an exponential equation.

$\displaystyle log_3(14x)=156$

To convert from logarithms to exponents, recall the following property:

$\displaystyle log_b(x)=a$

Can be rewritten as:

$\displaystyle b^a=x$

So, starting with

$\displaystyle log_3(14x)=156$,

We can get

$\displaystyle 3^{156}=14x$

To solve the following, you must "undo" the 5 with taking log based 5 of both sides. Thus,

$\displaystyle 5^x=125$

$\displaystyle \log_5{5^x}=\log_5{125}$

$\displaystyle x=\log_5{125}$

The right hand side can be simplified further, as 125 is a power of 5. Thus,

$\displaystyle x=\log_5{125}$

$\displaystyle x=\log_5{5^3}$

$\displaystyle x=3$

Apply the Product of Powers Property to rewrite the second expression:

$\displaystyle 2^{x}+ 2 ^{x+2} = 171$

$\displaystyle 2^{x}+ 2 ^{2} \cdot 2 ^{x} = 171$

$\displaystyle 1 \cdot 2^{x}+ 4 \cdot 2 ^{x} = 171$

Distribute out: 

$\displaystyle (1 + 4) \cdot 2 ^{x} = 171$

$\displaystyle 5 \cdot 2 ^{x} = 171$

Divide both sides by 5:

$\displaystyle \frac{5 \cdot 2 ^{x}}{5} = \frac{171}{5}$

$\displaystyle 2 ^{x} = 34.2$

Take the natural logarithm of both sides (and note that you can use common logarithms as well):

$\displaystyle \ln 2 ^{x} = \ln 34.2$

Apply a property of logarithms:

$\displaystyle x \ln 2 = \ln 34.2$

Divide by $\displaystyle \ln 2$ and evaluate:

$\displaystyle \frac{x \ln 2}{\ln2} = \frac{\ln 34.2}{\ln2}$

$\displaystyle x = \frac{3.5322 }{0.6931 }$

$\displaystyle x \approx 5.10$

$\displaystyle 9 = 3^{2}$, so rewrite the expression at right as a power of 3 using the Power of a Power Property:

$\displaystyle 3^{2x+ 1} = 9^{2x- 4 }$

$\displaystyle 3^{2x+ 1} = (3^{2} )^{2x- 4 }$

$\displaystyle 3^{2x+ 1} = 3^{2(2x- 4 )}$

Set the exponents equal to each other and solve the resulting linear equation:

$\displaystyle 2x+ 1 = 2(2x-4)$

Distribute:

$\displaystyle 2x+ 1 = 4x-8$

Subtract $\displaystyle 4x$ and 1 from both sides; we can do this simultaneously:

$\displaystyle 2x+ 1 - 4x - 1 = 4x-8 - 4x - 1$

$\displaystyle - 2x = -9$

Divide by $\displaystyle -2$:

$\displaystyle \frac{- 2x}{-2} = \frac{-9}{-2}$

$\displaystyle x =4.5$

To solve, you must first "undo" the log. Since no base is specified, you assume it is 10. Thus, we need to take 10 to both sides.

$\displaystyle 10^{\log(2x+4)}=10^1$

$\displaystyle 2x+4=10$

Now, simply solve for x.

$\displaystyle 2x=6$

$\displaystyle x=3$