Solve the equations 2(7^x = 3126, 5^(x-4) = 160.56

[jimmypop]

I just need the formula or method for solving something like this....

2(7^x)=3126
or
5^(x-4)=160.56

 

[Loren]

Take the log of both sides.

Example: Solve 2^(x+1)=5. Solve for x.

log 2^(x+1) = log 5
(x+1)(log 2) = log 5
x+1 =(log 5)/(log 2)
x = (log 5)/(log 2) - 1
x= (0.69897)/(0.30103) - 1
x = 2.32192 - 1 =1.32
Then check it.
2^2.32 =4.99

 

[jimmypop]

Thanks. That makes since for the second example, but I am still stuck on the first one 2(7^x)=3126? If the 2 wasn't there it would be simple, but that 2 is throwing me off?
I've tried to work it every way I can think of, but I can seem to get an answer that checks out?
Can I get any help on that one?

 

[mmm4444bot]

What would you do with the 2, if trying to solve the following?

2A = 3126

 

[jimmypop]

I would divide.

Here is what I tried....

log(3126)/log(3.5)=6.424
log(1563)/log(3.5)=5.871
log(3126)/log(7)=4.136 4.136/2=2.068


I know the answer is about 3.779395 I found that through trial and error. I just can't seem to do it quite the right way?

 

[jimmypop]

I've also tried (log(3126)\log(7))\log2 but that gave me 13.738
I've tried dividing the log of each separately by 2 then subtracting them also? I am just completely lost on this one?

 

[Loren]

log 2a[sup:liay18jb]3[/sup:liay18jb] = log 2 + log a[sup:liay18jb]3[/sup:liay18jb] = log 2 + 3log a.

 

[mmm4444bot]

I would divide …

Then do the same with 2(7^x) = 3126.

Do you know how to use the Change-of-Base Formula? Scientific calculators do not have a button for base-7 logarithms, so you need to use the formula to convert the log_7 to either log_10 or log_e.

 

[mmm4444bot]

… Here is what I tried …

… log(1563)/log(3.5) …

This looks like the expression we get, after using the Change-of-Base Formula, EXCEPT that the base is not 3.5; it's 7.

 

[jimmypop]

AHHH, I get it. I just Wasn't doing the order of operations right I guess?

So I can go like this......

2(7^x)=3126
7^x=3126/2
7^x=1563
log(1563)/log(7) which equals 3.779

so x would be "about" 3.779

Correct?

 

[mmm4444bot]

so x would be "about" 3.779

Correct?

I'll tell you that your work looks correct.

If you would like to confirm your numerical candidate, then stick the following into your machine, and see what you think. :wink:

2 * 7^3.779

If the result concerns you, then enter this: 2 * 7^3.77939

'

log(1563)/log(7)

I'm hoping that you wrote this through comprehension of the Change-of-Base Formula, and not simply changing the 3.5 to 7 (from what I wrote above). My point: Make sure that you understand how to apply the formula, and remember it.

'

… Wasn't doing the order of operations right I guess? …

I'm not sure about your thoughts because it seems like that 2 was flying all over the place. When solving an equation for a variable that appears in the exponent position, we need to first isolate the power containing it. Solving the following, for example, will require the same strategy: first isolate 7^x; then apply logarithms.

\(\displaystyle \frac{\sqrt{3 - 7^x}}{45} \; = \; 3126\)

 

[jimmypop]

Thanks, My father passed away a few weeks ago and I had to miss a couple classes. Obviously it was the classes that dealt with logs. Tomorrow morning is a test and This text book explains a few things, but no where near the examples I am looking at? It's driving me crazy. I've gotten all A's in this class, now I'm afraid that this test is going to bring down my average considerably..

Anyways, I don;t mean to ramble, but thanks again.