In order to simplify this expression, use the following natural log rule.

\(\displaystyle ln_x(x^y) =y\)

The natural log has a default base of \(\displaystyle e\). This means that: 

 \(\displaystyle 2e\cdot ln(e^3)=2e\cdot ln_e(e^3) = 2e\cdot (3)\)

The answer is:  \(\displaystyle 6e\)

According to log properties, the coefficient \(\displaystyle 4\) in front of the natural log can be rewritten as the exponent raised by the quantity inside the log.

\(\displaystyle 2e^{4ln{(2e)}} = 2e^{ln[{(2e)^4}]}\)

Notice that natural log has a base of \(\displaystyle e\).  This means that raising the log by base \(\displaystyle e\) will eliminate both the \(\displaystyle e\) and the natural log.

The terms become:  \(\displaystyle 2(2e)^4\)

Simplify the power.

\(\displaystyle 2(2e)^4 = 2(16e^4) = 32e^4\)

The answer is:  \(\displaystyle 32e^4\)

The natural log has a default base of \(\displaystyle e\).  Natural log to of an exponential raised to the power will be just the power.  The natural log and \(\displaystyle e\) will be eliminated.

\(\displaystyle ln(e^x)=x\)

Rewrite the expression.

\(\displaystyle 2e^2ln(e^2) = 2e^2(2) = 4e^2\)

The answer is:  \(\displaystyle 4e^2\)

The natural log has a default base of \(\displaystyle e\).  

According to the rule of logs, we can use:

\(\displaystyle log_{b}(b^x) = x\)

\(\displaystyle 6ln(e^{-3x}) = 6ln_e(e^{-3x})\)

The coefficient in front of the natural log can be transferred as the power of the exponent.

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

The natural log and base e will cancel, leaving just the exponent.

\(\displaystyle (-3x)(6)=-18x\)

The answer is:  \(\displaystyle -18x\)

The natural log has a default base of \(\displaystyle e\).  This means that the natural log of \(\displaystyle e\) to the certain power will be just the power itself.

\(\displaystyle log_{b}(b^x) = x\)

The expression \(\displaystyle 3ln(e^{7e})\) becomes:  

\(\displaystyle 3(7e) = 21e\)

The answer is:  \(\displaystyle 21e\)

Use the log properties to separate each term.  When the terms inside are multiplied, the logs can be added.

Rewrite the expression.

\(\displaystyle ln(6e^7) = ln(6)+ln(e^7) = ln(2)+ln(3)+ln(e^7)\)

The exponent, 7 can be dropped as the coefficient in front of the natural log.  Natural log of the exponential is equal to one since the natural log has a default base of \(\displaystyle e\).

\(\displaystyle log_{b}(b^x) = x\)

The answer is:  \(\displaystyle ln(2)+ln(3)+7\)

Use base \(\displaystyle e\) and raise both the left and right sides as the powers of \(\displaystyle e\).  This will eliminate the natural log term.

\(\displaystyle e^{kt} =e^ {ln(\frac{T-S}{T_0-S})}\)

The equation becomes:

\(\displaystyle e^{kt} =\frac{T-S}{T_0-S}\)

Multiply the quantity \(\displaystyle (T_0-S)\) on both sides.

\(\displaystyle e^{kt}(T_0-S) =\frac{T-S}{T_0-S}(T_0-S)\)

\(\displaystyle e^{kt}T_0-e^{kt}S =T-S\)

Add \(\displaystyle e^{kt}S\) on both sides.

\(\displaystyle e^{kt}T_0-e^{kt}S +e^{kt}S=T-S+e^{kt}S\)

The equation becomes:

\(\displaystyle e^{kt}T_0=T-S+e^{kt}S\)

To isolate \(\displaystyle T_0\), divide \(\displaystyle e^{kt}\) on both sides.

\(\displaystyle \frac{e^{kt}T_0}{e^{kt}}=\frac{T-S+e^{kt}S}{e^{kt}}\)

The answer is:  \(\displaystyle T_0 =\frac{T-S+e^{kt}S}{e^{kt}}\)

The natural log has a default base of \(\displaystyle e\).

Use the log property:

\(\displaystyle log_x(x^y)=y\)

We can cancel the base and the log of the base.

The expression \(\displaystyle 2e^2ln(e^{2e})\) becomes:

\(\displaystyle 2e^2(2e )= 4e^3\)

The answer is:  \(\displaystyle 4e^3\)

Notice that the terms inside the log are added, with a common factor of \(\displaystyle e\).  Pull out a common factor.

\(\displaystyle ln(e+e^2)= ln((e)(1+e))\)

Notice that these two terms inside the log are multiplied.  We can split the log into two terms.

\(\displaystyle ln((e)(1+e)) = ln(e)+ln(e+1)\)

The value of the first term is equal to one, since natural log has a default base of \(\displaystyle e\).  We can use the property \(\displaystyle ln_xx^y=y\) to eliminate the log and the \(\displaystyle e\) term, which will cancel leaving just the power of one.

The expression becomes:  \(\displaystyle 1+ln(e+1)\)

We cannot use the property of logs to simplify the second term.

The answer is:  \(\displaystyle 1+ln(e+1)\)

In order to eliminate the natural log and solve for x, we will need to exponential both sides because \(\displaystyle e\) is the base of natural log.

\(\displaystyle e^{ln(2x+e)}= e^{13}\)

The left side will be reduced to just the inner quantity of the natural log.

\(\displaystyle 2x+e = e^{13}\)

Subtract \(\displaystyle e\) from both sides of the equation.

\(\displaystyle 2x+e -e= e^{13}-e\)

\(\displaystyle 2x= e^{13}-e\)

Divide by two on both sides.

\(\displaystyle \frac{2x}{2}= \frac{e^{13}-e}{2}\)

The answer is:  \(\displaystyle \frac{e^{13}-e}{2}\)