logs. is my work correct?

• May 22nd 2007, 05:19 PM
trancefanatic
logs. is my work correct?
are these right?

3^2x=75 ---> i got log75/2log3

solve for x
log(2)x = 5 ---> i got x = 32

solve for x
e^3x-2=5 ----> i got x ~ 0.6486

solve for x
2^2x=8^x-4

2^2x=(2^3)x-4
2^2x=2^3x-4
2x+4=3x <--- as far as i could get
• May 22nd 2007, 05:37 PM
Jonboy
Hmm somebody check me here but I'm pretty confident.

Quote:

1) 3^2x=75 ---> i got log75/2log3
I didn't get that.

I'm assuming you mean: $3^{2x}\,=\,75$

Change to logarithmic form: $log_3\,75\,=\,2x$

You can get the answer now.

Quote:

2) log(2)x = 5 ---> i got x = 32
If you mean: $log_{2}\,x\,=\,5$ then yes.

Someone else help on 3.
• May 22nd 2007, 05:48 PM
Jhevon
Quote:

Originally Posted by trancefanatic

solve for x
2^2x=8^x-4

2^2x=(2^3)x-4
2^2x=2^3x-4
2x+4=3x <--- as far as i could get

all were fine except for the above. i don't see how you got stuck at that last line though, it's simple algebra from there. to be able to solve logarithmic equations and not be able to solve that is like knowing how to run without knowing how to walk, but anyway, let's see where you made the mistake.

first of all, is the -4 a part of the power of 8 or is it separate? i will do both, choose which is right. Please make sure to type your questions clearly next time.

$2^{2x} = 8^{x-4}$

$\Rightarrow 2^{2x} = (2^3)^{x-4}$

$\Rightarrow 2^{2x} = 2^{3x - 12}$

equating the powers we get:

$2x = 3x - 12$

$\Rightarrow 0 = x - 12$ .........subtracted $2x$ from both sides

$\Rightarrow x = 12$

Or, if the power of 8 was actually x, and the -4 was separate, the problem becomes a bit more interesting
$2^{2x} = 8^{x} - 4$

$\Rightarrow 2^{2x} = 2^{3x} - 4$

Let $y = 2^x$, we get:

$y^2 = y^3 - 4$

$\Rightarrow y^3 - y^2 - 4 = 0$

we see that $y = 2$ is a root, so $y - 2$ is a factor.
Dividing the equations by $y - 2$ using long division (or synthetic division) we obtain:

$y^3 - y^2 - 4 = (y - 2)(y^2 + y + 2) = 0$

$\Rightarrow y - 2 = 0 \mbox{ or } y^2 + y + 2 = 0$

only $y - 2 = 0$ has real roots, so

$y = 2$

But $y = 2^x$

$\Rightarrow 2^x = 2$

$\Rightarrow x = 1$ by equating the coefficients
• May 22nd 2007, 05:50 PM
Jhevon
Quote:

Originally Posted by Jonboy
Hmm somebody check me here but I'm pretty confident.

I didn't get that.

I'm assuming you mean: $3^{2x}\,=\,75$

Change to logarithmic form: $log_3\,75\,=\,2x$

You are also right. the difference is he used $\log_{10}$, it makes the answer a bit messier, but it's right nonetheless
• May 22nd 2007, 05:59 PM
trancefanatic
Quote:

Originally Posted by Jhevon
all were fine except for the above. i don't see how you got stuck at that last line though, it's simple algebra from there. to be able to solve logarithmic equations and not be able to solve that is like knowing how to run without knowing how to walk, but anyway, let's see where you made the mistake.

i admit that i got lazy and didn't see the next steps. baby steps i know. thats just me makin it harder then it really is. but thank you.
• May 22nd 2007, 06:03 PM
Jhevon
Quote:

Originally Posted by trancefanatic
i admit that i got lazy and didn't see the next steps. baby steps i know. thats just me makin it harder then it really is. but thank you.

what was the actual power of 8? was it just x, or was it x - 4 ?
• May 22nd 2007, 06:07 PM
trancefanatic
Quote:

Originally Posted by Jhevon
what was the actual power of 8? was it just x, or was it x - 4 ?

the power was x-4.
• May 22nd 2007, 06:08 PM
Jhevon
Quote:

Originally Posted by trancefanatic
the power was x-4.

ok, so you want the first problem i did. do you see your mistake? you did not distribute the 3 correctly. you multiplied the x by 3 but not the - 4, this made your answer incorrect

also you equatied the powers incorrectly
• May 22nd 2007, 06:13 PM
trancefanatic
Quote:

Originally Posted by Jhevon
ok, so you want the first problem i did. do you see your mistake? you did not distribute the 3 correctly. you multiplied the x by 3 but not the - 4, this made your answer incorrect

also you equatied the powers incorrectly

yeah, i realized i needed to distribute it to the whole thing. thanks