Please Help: Exponential and Log. functions.

E

edd fedds

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Joined
Dec 11, 2009
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Gott stuck and/or confused on the solution for a couple problems. Please help me. Below are the problems and my attempta at a solution.

1.)Solve the equation log(subscript 3)(3x-6)=log(subscrpt 3)(2x+1).
a.-1
b.1
c.3
d.7
*For this problem i did:
log(sub. 3)=log(sub 3)(3x-6)-log(sub. 3)(2x+1)
log(sub. 3)=log(sub 3)(3x-6)/(2x+1)
x=1.14. So I chose B. Am I right?

2.)Solve the equation 3log(subcript 5)x-log(subcript 5)=log(subcript 5)16.
a.4
b.8
c.12
d.16
*For this problem i did:
3log(subcript 5)x-log(subcript 5)=log(subcript 5)16
log(sub. 5)x^3=log(sub. 5)16/log(sub 5)4
log(sub. 5)x^3=log(sub. 5)4
x^3=4
x=64
x=(squareroot of)64=8. So I chose B. Am I right?

3.)Solve the equation 1/3log(subscript 7)64+1/2log(subscript 7)121=log(subscript 7)x.
a.7
b.15
c.19
d.44
*For this problem I did:
1/3log(subscript 7)64+1/2log(subscript 7)121=log(subscript 7)x
log(sub. 7)64^1/3+log(sub. 7)121^1/2=log(sub. 7)x
log(sub. 7)4+log(sub. 7)11=log(sub. 7)x
log(sub. 7)(11)(4)=log(sub. 7)x
44=x. So I chose D. Am I right? Should I Square root it?

4.)Solve the equation 100e^.75=500. Round the answer to four decimal places.
a.2.1458
b.3.1266
c.4.2134
d.5.5945
*For this problem I did:
100e^.75=500
ln500=100e^.75x
ln500=ln100+.75(1)
6.2146=ln100.75
=1.3400. Unsure of why I gott this answer.

5.) Solve the equation 6^(3x+1)=8. Round the answer to four decimal places.
a.) .0021
b.) .0535
c.) .0721
d.) .0839
* Need help period. My answer was -4.2219. Please help. THANK YOU. :)
 
edd fedds said:
Gott stuck and/or confused on the solution for a couple problems. Please help me. Below are the problems and my attempta at a solution.

1.)Solve the equation log(subscript 3)(3x-6)=log(subscrpt 3)(2x+1).
a.-1
b.1
c.3
d.7
*For this problem i did:
log(sub. 3)=log(sub 3)(3x-6)-log(sub. 3)(2x+1)
log(sub. 3)=log(sub 3)(3x-6)/(2x+1)
x=1.14. So I chose B. Am I right?
Click to expand...

log[sub:329iuzme]3[/sub:329iuzme] (3x - 6) = log[sub:329iuzme]3[/sub:329iuzme] (2x + 1)

Since the bases are the same, and the logs are equal, the arguments must be equal as well...

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

Solve that for x.

2.)Solve the equation 3log(subcript 5)x-log(subcript 5)=log(subcript 5)16.
a.4
b.8
c.12
d.16
*For this problem i did:
3log(subcript 5)x-log(subcript 5)=log(subcript 5)16
log(sub. 5)x^3=log(sub. 5)16/log(sub 5)4
log(sub. 5)x^3=log(sub. 5)4
x^3=4
x=64
x=(squareroot of)64=8. So I chose B. Am I right?
Click to expand...

Please check to make sure you've typed the problem correctly.

3 log[sub:329iuzme]5[/sub:329iuzme] x - log[sub:329iuzme]5[/sub:329iuzme] ?????<----there is no argument for this log....log[sub:329iuzme]5[/sub:329iuzme] WHAT???? = log[sub:329iuzme]5[/sub:329iuzme] 16

As stated, the problem makes no sense.

3.)Solve the equation 1/3log(subscript 7)64+1/2log(subscript 7)121=log(subscript 7)x.
a.7
b.15
c.19
d.44
*For this problem I did:
1/3log(subscript 7)64+1/2log(subscript 7)121=log(subscript 7)x
log(sub. 7)64^1/3+log(sub. 7)121^1/2=log(sub. 7)x
log(sub. 7)4+log(sub. 7)11=log(sub. 7)x
log(sub. 7)(11)(4)=log(sub. 7)x
44=x. So I chose D. Am I right?Yes Should I Square root it?No! Why would you want to do that?

[quote:329iuzme]
4.)Solve the equation 100e^.75=500. Round the answer to four decimal places.
a.2.1458
b.3.1266
c.4.2134
d.5.5945
*For this problem I did:
100e^.75=500
ln500=100e^.75x<------where did the x come from? There's no "x" in the original problem
ln500=ln100+.75(1)
6.2146=ln100.75
=1.3400. Unsure of why I gott this answer.
Click to expand...

Again, you need to check to see if you've typed the problem correctly. 100 e[sup:329iuzme].75[/sup:329iuzme] = 500 does not have a variable in it. And 100 * e[sup:329iuzme].75[/sup:329iuzme], according to my calculator, is 211.7000017, not 500

If, perhaps, the problem was 100 e[sup:329iuzme].75x[/sup:329iuzme] = 500, then you need to start by dividing both sides by 100:

e[sup:329iuzme].75x[/sup:329iuzme] = 5

Then, take the natural log of both sides:

ln e[sup:329iuzme].75x[/sup:329iuzme] = ln 5
.75x = ln 5

And solve for x.

5.) Solve the equation 6^(3x+1)=8. Round the answer to four decimal places.
a.) .0021
b.) .0535
c.) .0721
d.) .0839
* Need help period. My answer was -4.2219. Please help. THANK YOU. :)
Click to expand...
[/quote:329iuzme]

Take the natural log of both sides:

ln 6[sup:329iuzme]3x + 1[/sup:329iuzme] = ln 8

(3x + 1)*ln 6 = ln 8

3x * ln 6 + 1*ln 6 = ln 8

3x * ln 6 = ln 8 - ln 6

(3 ln 6)*x = ln 8 - ln 6

Divide both sides by 3 ln 6:

x = (ln 8 - ln 6) / (3 ln 6)

Looks like a good review of the rules of logs might be in order.
 
Thank you so much for your help. I'm still a little stuck on number one. Could go about breaking that down for me.Alos i made a typo in problem two. I meant Log(subscript 5)4 not log(subscript 5) alone. Could someone please help me on number one and two. Please and Thank you.
 
edd fedds said:
Thank you so much for your help.

I'm still a little stuck on number one. Exactly where are you lost? Could go about breaking that down for me.Alos i made a typo in problem two.


I meant Log(subscript 5)4 not log(subscript 5) alone. Could someone please help me on number one and two. Please and Thank you.
Click to expand...
\(\displaystyle 3log_5x - log_54 = log_516\)

\(\displaystyle 3log_5x = log_54 + log_516\)

\(\displaystyle log_5x^3 = log_564\)

\(\displaystyle log_5x^3 = log_54^3\)

Now continue...

Like Mrs? said - a review of "Logs" is in order...
 
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