Solve(Using Logarithms)

Question

Another problem I can't quite figure out:

${1\over2}\ln(a+1) + \ln 5 = 1$

Solve using logarithms.

Thank you!

Answer 1

You need to use the following identity: $$\ln a + \ln b = \ln ab$$ $$\ln (a+1)^{\frac{1}{2}} + \ln 5 = 1$$ $$\ln 5(a+1)^{\frac{1}{2}} = 1$$ $$5(a+1)^{\frac{1}{2}} = e^1$$ $$25(a+1) = e^2$$ $$a = \frac{e^2}{25}-1$$

Answer 2

Using $m\log(c)=\log(c^m)$ where both logarithm remain defined

$$\implies \ln(a+1)+2\ln 5=2\iff\ln(a+1)+\ln(5^2)=2$$

Again, using $\log(x)+\log(y)=\log(xy)$ where both logarithm remain defined

$$\iff\ln\{25(a+1)\}=2$$

$$\implies25(a+1)=e^2$$

Answer 3

Would be pretty hard to solve it without logarithms as they are given in the equation, but fortunately we are even asked to use them. First receipt in problems of that kind that look hard to you: introduce a new variable in which the original equation looks better. For example, let's say $b = \ln(a+1)$, then $\frac12b = 1-\ln5$ and so that $b = 2-\ln25$, so you get $\ln(a+1)=2-\ln25$. All is left is to exponentiate both sides.