# Questions about exponents and absolute values

• Sep 6th 2009, 03:51 PM
ktpinnock
Questions about exponents and absolute values
Ok the first one is: 7^2x+3 = 5^3x-2

I Understand that I can put them in log form, (2x+3)log 7 = (5x-2)log5
but, my teacher mentioned we should use the natural log? So how do I go about doing so?

2. 8^x2 * 4^-3x = 1/2
How I did it:
2^(3x^2) * 2^-6x = 1/2
= 3x^2 * -6x = 1/2
= -18x^3 = 1/2

Some how that doesn't seem right to me.
I thought I could make the bases equal, so I used 2 for both the 8 and the 4. is this correct? does the 1/2 have to have to the same base also?

3. 8-2|6-2x| ≤ -4
-8 -8
= -2|6-2x| ≤ -12

then I divided by 2 on both sides to get |6-2x| ≤ -12

Is that right? should I be dividing by -2 or adding?

4. Last one:
2^x+1 * 8^-x = 4

Same as the first one, I set the bases as 2, So i get
2^x+1 * 2^-3x = 2^2
Should i be doing that for the 4 also? it doesn't seem right again...Thx in advance for the help.
• Sep 6th 2009, 04:09 PM
skeeter
Quote:

Originally Posted by ktpinnock
Ok the first one is: 7^2x+3 = 5^3x-2

I Understand that I can put them in log form, (2x+3)log 7 = (5x-2)log5
but, my teacher mentioned we should use the natural log? So how do I go about doing so?

doesn't matter ... same a log base 10.

(2x+3)ln 7 = (3x-2)ln 5

solve for x

2. 8^x2 * 4^-3x = 1/2
How I did it:
2^(3x^2) * 2^-6x = 1/2

2^(3x^2-6x) = 2^(-1)

3x^2 - 6x = -1

3. 8-2|6-2x| ≤ -4
-8 -8
-2|6-2x| ≤ -12

divide both sides by -2 ... dividing by a negative value changes the direction of the inequality.

|6-2x| > 6

4. Last one:
2^x+1 * 8^-x = 4

2^(x+1) * 2^(-3x) = 2^2

2^(1-2x) = 2^2

exponents are equal ...

...
• Sep 6th 2009, 04:36 PM
ktpinnock
o0o0o, thanks a lot for the quick response and clearing the mist that was forming over my brain (Rock)

One more quick question...
(4 is the base)
log4 (x) + log 4 (x-4) = 1, Solve for x:

I did log4 (x)(x-4) = 1

i get confused after that because my teacher said it should be set = 4..which I don't get, can you please explain? or anyone for that matter...thanks again
• Sep 6th 2009, 04:46 PM
skeeter
Quote:

Originally Posted by ktpinnock
o0o0o, thanks a lot for the quick response and clearing the mist that was forming over my brain (Rock)

One more quick question...
(4 is the base)
log4 (x) + log 4 (x-4) = 1, Solve for x:

I did log4 (x)(x-4) = 1

i get confused after that because my teacher said it should be set = 4..which I don't get, can you please explain? or anyone for that matter...thanks again

one method ...

\$\displaystyle \log_4[x(x-4)] = 1\$

\$\displaystyle \log_4[x(x-4)] = \log_4{4}\$

\$\displaystyle x(x-4) = 4\$

another ...

\$\displaystyle \log_4[x(x-4)] = 1\$

change to an exponential ...

\$\displaystyle 4^1 = x(x-4)\$

now solve ... don't forget to check any solution(s) in the original equation.
• Sep 6th 2009, 05:17 PM
ktpinnock
ooo ok, I was on the right track...thanks for the directions..I see ur also a fellow spammer of ellipses (Giggle)

Edit: Last one I swear D:

log(7x-12) = 2logx

Can i just drop the log since they are the same base?

and write 7x-12 = x^2?
• Sep 6th 2009, 05:48 PM
skeeter
Quote:

Originally Posted by ktpinnock
ooo ok, I was on the right track...thanks for the directions..I see ur also a fellow spammer of ellipses (Giggle)

correction, ... ellipsis

ellipses ...

• Sep 6th 2009, 06:00 PM
ktpinnock
Quote:

Originally Posted by skeeter
correction, ... ellipsis

ellipses ...

Actually ellipses is also the plural of ellipsis, which is what i was going for...but i guess this IS a math forum...
• Sep 6th 2009, 06:08 PM
skeeter
Quote:

Originally Posted by ktpinnock
Actually ellipses is also the plural of ellipsis ...

... and you learn something new every day.
• Sep 6th 2009, 06:20 PM
ktpinnock
Quote:

Originally Posted by skeeter
... and you learn something new every day.

yea aha...well since your still here...

log(7x-12) = 2logx

Can i just drop the log since they are the same base?

and write 7x-12 = x^2?

and Factoring..

64a^6 + 27b^3
i did (8a^3) (8a^3) + (3b)(3b)(3b)

and for 2x^4-162
2(x^4 - 81)
2(x^2 - 9) (x^2 +9)
2(x - 3) (x+3) (x^2 - 9)

Right? (doubting they are tho)
• Sep 6th 2009, 06:29 PM
skeeter
Quote:

Originally Posted by ktpinnock
yea aha...well since your still here...

log(7x-12) = 2logx

Can i just drop the log since they are the same base?

and write 7x-12 = x^2? yes

and Factoring..

64a^6 + 27b^3
i did (8a^3) (8a^3) + (3b)(3b)(3b)

(4a^2)^3 + (3b)^3

factoring pattern is x^3 + y^3 = (x + y)(x^2 - xy + y^2)

and for 2x^4-162
2(x^4 - 81)
2(x^2 - 9) (x^2 +9)
2(x - 3) (x+3) (x^2 + 9)

btw ... start new problems with a new thread.
• Sep 6th 2009, 06:41 PM
ktpinnock
Oh okay, I figured I'd just do it here instead of wasting thread space, but I'll do that next time, and thanks for being patient with my slowness.