Help with logarithms pls

1. May 22, 2005

Nx2

hi guys, im not too sure how to do these questions:
solve for x,
a) (logx^3)^2 = logx^18
b) logx^3 + log(x^logx) = -2

so this is what i got so far:
a) (3logx)^2 = logx^18
9logx^2 = logx^18
18logx = logx^18
logx^18 = logx^18
... then i got stuck.... i was clueless and did not know how to solve for x.

b) 3logx + (logx)(logx) = -2
logx^2 + 3logx +2 = 0
so let logx = s
s^2 +3s + 2 = 0
(s+1)(s+2)=0
s=-1 or s=-2
... then again i got stuck... im not sure if what i did was right. our teacher gave us these questions that she has never taught us b4 and said this is ur assignment for the break... any help would be very much appreciated. thnx.

- Tu

2. May 22, 2005

quasar987

a) You did an illegal operation btw line 2 and line 3 when you said 9logx^2 = 18logx

What you got here is $9(logx)^2$ and not $9log(x^2)$. And while it is true that $9log(x^2)=18logx$, it is not that $9(logx)^2=18logx$

3. May 22, 2005

quasar987

b) You're almost there! you got the solutions s=-1 and s=-2, which translate into logx =-1 and logx=-2. Now you gotta solve for x. Hint: Use the fact that $e^x$ is the inverse function of $logx$. This means that

$$e^{logx}=x$$

mmh.

4. May 22, 2005

Nx2

oo... ok... so now that i have that... how do i solve for x in a though?

5. May 22, 2005

quasar987

see my last post. same trick.

6. May 22, 2005

Nx2

srry.. but i dont think we learned e^x

7. May 22, 2005

quasar987

Mmmh, did you learn about exp(x) ?

Last edited: May 22, 2005
8. May 22, 2005

Nx2

yea i know exp(x)

9. May 22, 2005

quasar987

They are the same.

$$\mbox{exp}(x) = e^x$$

So you know that exp(x) is the inverse function of logx, right?

Then exp(logx)=x.

10. May 22, 2005

Nx2

ok.. but i still dont get how i would solve for x if i had
logx^18 = logx^18... wouldnt i just end up with 0 = 0?

11. May 22, 2005

quasar987

err. you don't have logx^18 = logx^18. I told you there was a mistake between line 2 and 3. All you have is

$$9(logx)^2 = (logx)^{18}$$

If you don't see where to go from there, use the same substitution as you did in b). Set s = logx... then solve for s. Then use the fact that exp(s)=x to solve for x.

Last edited: May 22, 2005
12. May 22, 2005

Nx2

omg... thats why it wasnt working.... i was like so clueless... forgot bout that... thnx alot man. i c what u meant when u were talking bout e^x now... thnx, very much appreciated for putting up with me.

13. May 22, 2005

quasar987

It's all good. Don't hesistate to post if something doesn't work out!