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

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- May 22nd 2007, 05:19 PMtrancefanaticlogs. 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 PMJonboy
Hmm somebody check me here but I'm pretty confident.

Quote:

1) 3^2x=75 ---> i got log75/2log3

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

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

You can get the answer now.

Quote:

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

Someone else help on 3. - May 22nd 2007, 05:48 PMJhevon
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.

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

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

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

equating the powers we get:

$\displaystyle 2x = 3x - 12$

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

$\displaystyle \Rightarrow x = 12$

Or, if the power of 8 was actually x, and the -4 was separate, the problem becomes a bit more interesting

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

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

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

$\displaystyle y^2 = y^3 - 4$

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

we see that $\displaystyle y = 2$ is a root, so $\displaystyle y - 2$ is a factor.

Dividing the equations by $\displaystyle y - 2$ using long division (or synthetic division) we obtain:

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

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

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

$\displaystyle y = 2$

But $\displaystyle y = 2^x$

$\displaystyle \Rightarrow 2^x = 2$

$\displaystyle \Rightarrow x = 1$ by equating the coefficients - May 22nd 2007, 05:50 PMJhevon
- May 22nd 2007, 05:59 PMtrancefanatic
- May 22nd 2007, 06:03 PMJhevon
- May 22nd 2007, 06:07 PMtrancefanatic
- May 22nd 2007, 06:08 PMJhevon
- May 22nd 2007, 06:13 PMtrancefanatic