logarithmic questions

[Anood]

Can you please help me in sloving these 3 problems:

solve for x:

1)e^(2x)- 7 e^(x) +12=0

2)3(e)^(x) - 5= -2 e^(-x)

3) 2lnx- 3 ln (1/x)=10

Here is my attempt:
1) e^(2x)- 7 e^(x) +12=0
ln e^(2x)- ln 7 e^(x) + ln 12= ln 0
2x-7 ln e^(x)+ln 12=1
2x-7x+ln 12=1
-5x+ln 12=1

2)3(e)^(x) - 5= -2 e^(-x)
ln 3(e)^(x) - ln 5= ln -2 e^(-x)
3ln (e)^(x) - ln 5= -2 ln e^(-x)
3x-ln5=-2(-x)

3)i don't know how to do it

 

[galactus]

For the first one, a u substitution is a good way to go.

\(\displaystyle e^{2x}-7e^{x}+12=0\)

Notice it is in the form of a quadratic.

Let \(\displaystyle u=e^{x}\)

Then you have:

\(\displaystyle u^{2}-7u+12=0\)

Factor:

\(\displaystyle (u-4)(u-3)=0\)

\(\displaystyle u=4, \;\ u=3\)

Therefore, \(\displaystyle 4=e^{x}, \;\ 3=e^{x}\)

Solve for x:

\(\displaystyle x=2ln(2), \;\ x=ln(3)\)

See how that works?. Try using it with other problems if the need arises.

 

[soroban]

Hello, Anood!

Your steps are illegal . . .

\(\displaystyle 1)\;\;e^{2x}- 7 e^x +12\:=\:0\)

Here is my attempt:

\(\displaystyle e^{2x}- 7e^x +12\:=\:0\)

\(\displaystyle \ln\left(e^{2x}\right)- \ln\left(7e^x\right) + \ln(12)\:= \:\ln(0)\) . . . . NO!
Besides, ln(0) does not exist.

Think of it as a quadratic: .\(\displaystyle (e^x)^2 - 7(e^x) + 12 \:=\:0\)

. . and it factors: .\(\displaystyle (e^x - 3)(e^x - 4) \:=\:0\)


Can you finish it now?

 

[Anood]

Thanks a lot i finished up the first 2 but stuck in the third one.

Using the quotient rule
2ln(x)-3ln(1/x)=ln(x^2)+ln(x^3)=ln(x^5)

ln(x^5)=10
then what should i do?

 

[galactus]

The third one.

\(\displaystyle 2ln(x)-3ln(\frac{1}{x})=10\)

Use the law of logs, \(\displaystyle ln(\frac{a}{b})=ln(a)-ln(b)\)

\(\displaystyle 2ln(x)-3[\underbrace{ln(1)-ln(x)}_{\text{ln(1/x)}}]=10\)

Now, can you finish?.

 

[Deleted member 4993]

\(\displaystyle \frac{1}{x} = x^{-1}\)

so

\(\displaystyle ln(\frac{1}{x}) = ln(x^{-1})\)

then

\(\displaystyle ln(\frac{1}{x}) = - ln(x)\)

Now continue...