How to solve fancy natural log problems

Question

These are three 'fancy' natural log problems my teacher gave. I call them fancy because of how he set them up.

1.$$10^x=e^{2x-1}$$

2.$$\ln(x+5)-\ln(x-5)=4$$

3.$$\ln(4-x^2)+\ln 4=\ln15$$

For the first one this is how I did it:

$$x\ln10=(2x-1)\ln(e)$$ $$x=(2x-1)\ln(e)/\ln(10)$$ $$-x=-1(1/\ln10)$$ $$ x=1/(1/\ln10) \approx 2.303$$

The second one: $$\ln(x+5÷x-5)=4$$ $$\ln(x+5)=4x-20$$ $$\ln(x)=4x-15$$ I stopped there.

The third one: $$\ln(4-x^2)+\ln4=\ln15$$ $$\ln(16-4x^2)=\ln15$$ $$4x^2=\ln15+\ln16$$ $$x=\sqrt {\ln15+\ln16}/\sqrt{4}$$ $$x\approx 1.17$$

My teacher marked me wrong for all three problems. I've been trying my best to solve the question but I keep getting wrong answers.

Answer 1

Well known properties: $P1)~ \log(xy)=\log x+\log y,~~P2)\log(x/y)=\log x-\log y$ for $x,~y>0$. Also $P3)~\ln e^x=x$ for all $x$ and $P4)~$ $e^{\ln x}=x$ for $~x>0$.

About the second equation: $\ln (x+5)-\ln(x-5)=4$, equivalently $\displaystyle \ln\frac{x+5}{x-5}=4$ from $P2$, equivalently $\displaystyle \frac{x+5}{x-5}=e^4$ from $P4$. So $x+5=e^4(x-5)$, equivalently $\displaystyle x=5\frac{e^4+1}{e^4-1}.$

About the first one, your first line is not correct. About the third one, your third line is not correct, it shoud have been $16-4x^2=15$. The details are left for you to fill them up.

Answer 2

I give you some equiavalent equations from there I think you can go further. The first equation maybe is

1.$$10^x=e^{2x-1}$$ $$x\ln10=2x-1$$

2.$$\ln(x+5)-\ln(x-5)=4$$ $$\ln\frac{x+5}{x-5}=4$$ $$e^4=\frac{x+5}{x-5}$$ 3.$$\ln(4-x^2)+\ln 4=\ln15$$ $$\ln(4-x^2)=\ln15-\ln4$$ $$4-x^2=\frac{15}{4}$$

Answer 3

Let us solve the third equation

$ln(4-x^2)+ln(4)=ln(15)$.

at first, we must have $4-x^2>0$ or $-2<x<2$.

$ln(4(4-x^2))=ln(15)$

we take of $ln$ from both sides

$16-4x^2=15$

$4x^2=1$

$x=\frac{1}{2}$ or $x=\frac{-1}{2}$.

Answer 4

The third problem simplifies to:

$\ln[(4-x^2) 4] = \ln(15)$

$4(4-x^2) = 15$

$16 - 4x^2 = 15$

$x^2 = \frac{1}{4}$

The two solution are: $x_1 = \frac{1}{2}$ and $x_2 = -\frac{1}{2}$