Question

solve.
1. ln x+ln(x-2)=1
2. 3+ ln x+8
3. 3e^(-x)-4=9

Answer 1

first one
1) write as a single log via
\[\ln(x)+\ln(x-2)=\ln(x(x+2))\]
2) rewrite
\[\ln(x(x+2))=1\] in equivalent exponential form
\[x(x-2)=e\] and solve the resulting quadratic equation.
you will need to use the quadratic formula

Answer 2

second one is not an equation

Answer 3

for number 2 it's equals 8, not plus.

Answer 4

third one
1) add 4
2) divide by 3
3) write in equivalent logarithmic form
4) change the sing

Answer 5

i will write it if you like, but those are the steps

Answer 6

I can solve the first one for you:
Rule of logs:
\[loga+logb=log(ab)\]
So we have:
\[ln(x)(x-2)=1\]
Raise both sides by e^x:
\[x^2-2x=e\]
You can complete the squares or use quadratic formula to get:
\[1\pm\sqrt{1+e}\]

Answer 7

oh in that case we do it in two steps
1)
\[\ln(x)=5\]
\[x=e^5\]

Answer 8

For all of these the key is to realize that \[e^{ln(x)}=x\]

Answer 9

@osirisis yes of course you are right completing the square is much easier because you have
\[x^2-2x=e\] and -2 is even

Answer 10

so @Aggie forget i mentioned the quadratic formula

Answer 11

Also worth noting that ln can be used to eliminate e too.
\[ln (e^x)=x\]

Answer 12

osiris has written exactly what you need. also you should be able to switch from
\[\log_b(x)=y\iff b^y=x\] with ease
so in this case
\[\ln(x)=5 \iff e^5=x\] is immediate

Answer 13

So for the last porblem:
\[3e^{-x}-4=9\]
\[3e^{-x}=13\]
\[e^{-x}=\frac{13}{3}\]
\[-x=ln\frac{13}{3}\]
\[x=-ln\frac{13}{3}\]

Answer 14

Logs can look intimidating but if you approach them in a step by step process like normal problems they're not as bad as they seem.

Answer 15

exactly! and the steps were
1) add 4
2) divide by 3
3) write in equivalent logarithmic form
4) change the sign

Answer 16

so only step 3) had anything to do with exponentials and logs. the rest was elementary algebra