1. ## 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.

2. 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 ...

...

3. o0o0o, thanks a lot for the quick response and clearing the mist that was forming over my brain

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

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

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 ...

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

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

$x(x-4) = 4$

another ...

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

change to an exponential ...

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

now solve ... don't forget to check any solution(s) in the original equation.

5. ooo ok, I was on the right track...thanks for the directions..I see ur also a fellow spammer of ellipses

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?

6. 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
correction, ... ellipsis

ellipses ...

7. 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...

8. Originally Posted by ktpinnock
Actually ellipses is also the plural of ellipsis ...
... and you learn something new every day.

9. 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)

10. 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.

11. 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.