Simplifying $ \log_{e^2}(e^{4a}+ae^{4a}) $

Question

Problem

Simplify logaritm:

$$ \log_{e^2}(e^{4a}+ae^{4a}) $$

preferably in a way that end result contains only natural logarithm.

Attempt to solve

I know few computational rules about logarithms:

$$ \log_a(xy) = \log_a(x)+\log_a(y) $$

$$ \log_a(\frac{x}{y})=\log_a{x}-\log_a(y) $$

$$ \log_a(x^n)=n\log_a(x) $$

And formula for change of basis :

$$ \log_a(x)=\frac{\log_b(x)}{\log_b(a)} $$

There is sum inside this logarithm and it appears we don't have formula for this. On wikipedia i found two formulas about summation / substraction inside logarithm:

$$ \log_b(a+c)=\log_b a+ \log_b (1+\frac{c}{a}) $$

$$ \log_b(a-c)=\log_b a+ \log_b (1-\frac{c}{a}) $$

I have no former experience / knowledge of these formulas but for now the assumption is these are correct.

I would try to first try change of basis to $\log_{e^2}()\rightarrow \log_e() = \ln()$

$$ \log_{e^2}(e^{4a}+ae^{4a})=\frac{\log_e(e^{4a}+ae^{4a})}{\log_e(e ^2)} $$

$$ \log_{e}(e^{4a}+ae^{4a})=\log_{e^2}(e^{4a}+ae^{4a})\cdot \log_e({e^2}) $$

$$ \log_{e}(e^{4a}+ae^{4a})=2 \cdot \log_{e^2}(e^{4a}+ae^{4a})$$

We get the original logarithm with base $e$

$$ \log_{e^2}(e^{4a}+ae^{4a})=\frac{1}{2}\ln(e^{4a}+ae^{4a}) $$

Now the sum:

$$ \frac{1}{2}\ln((a+1)e^{4a}) $$

$$4a \frac{1}{2} \ln ((a+1)e) $$ $$ 4a \cdot \frac{1}{2} \ln (a+1) + \ln(e) $$

$$ \frac{4a}{2} \ln {(a+1)} + 1 $$

$$ 2a \ln (a+1) + 1 $$

If there is an error let me know.

Answer 1

Your answer is good, except for a slip in the second equation after your words “Now the sum”, because the exponent $4a$ does not apply to the factor $(1+a)$. But as usual when we’re beginners, there’s often a shorter, clearer way.

There are two questions here, the first of which is how to handle the unorthodox $\log_{e^2}$, while the second is how to handle the sum inside the parentheses.

For the first question, I recommend not depending on formulas that may seem to come from on high, but rather going back to the definitions: \begin{align} \log_b(x)=L\qquad&\text{means}\qquad b^L=x\\ \log_{e^2}(x)=Y\qquad&\text{means}\qquad (e^2)^Y=x&\text{so}\qquad e^{2Y}=x\\ \log_e(x)=2Y\qquad&\text{means}\qquad e^{2Y}=x&\text{so}\qquad Y={\scriptstyle\frac12}\log_e(x)\,. \end{align} That solves the first question: $\log_{e^2}(e^{4a}+ae^{4a})=\frac12\ln(e^{4a}+ae^{4a})$.

The second question now becomes easy: again, don’t rely on that funny rule from Wikipedia, but just factor the $e^{4a}$ from the two terms: $\frac12\ln(e^{4a}+ae^{4a})=\frac12\ln\bigl(e^{4a}(1+a)\bigr) =\frac12\bigl(\ln(e^{4a})+\ln(1+a)\bigr)$. Now remember that $\ln(e^{4a})=4a$ and get your answer $2a+\frac12\ln(1+a)$.

Answer 2

You were doing fine down to $$\frac 12\ln((a+1)e^{4a})$$ but then you cannot pull out the $4a$ because it is not an exponent on $a+1$. Instead you should do $$\frac 12\ln((a+1)e^{4a})=\frac 12\left(\ln (a+1)+\ln(e^{4a})\right)\\=\frac 12\ln(a+1)+2a$$

Answer 3

Factorizing the numerator and using change of basis :

$$\frac{\ln ((a+1)e^{4a})}{\ln e^2} =\frac 12 (\ln{(a+1)}+\ln e^{4a})$$ $$= \frac 12\ln{(a+1)} +2a$$

Answer 4

Using $$\log_{a}(x) = \frac{\ln(x)}{\ln(a)},$$ where $\ln(x) = \log_{e}(x)$, then \begin{align} \log_{e^{2}}(e^{4 a} + a e^{4 a}) &= \frac{\ln(e^{4 a}(1+a))}{\ln(e^{2})} \\ &= \frac{4 a + \ln(1 + a)}{2 \, \ln(e)} \\ &= 2 a + \frac{\ln(1+a)}{2}. \end{align}

Since $$ \frac{\ln(1 +a)}{2} = \frac{\ln(1+a)}{2 \, \ln(e)} = \frac{\ln(1+a)}{\ln(e^{2})} = \log_{e^{2}}(1+a),$$ then $$\ln_{e^{2}}(e^{4 a} + a e^{4 a}) = 2a + \log_{e^{2}}(1+a).$$

Answer 5

$$\log_{e^2}(e^{4a}+ae^{4a})$$$$=\log_{e^2}e^{4a}(1+a)$$$$=\log_{e^2}e^{4a}+log_{e^2}(1+a)$$$$=\log_{e^2}e^{2.2a}+log_{e^2}(1+a)$$$$=\log_{e^2}(e^{2})^{2a}+log_{e^2}(1+a)$$$$={2a}+log_{e^2}(1+a)$$