3 Logarithm Problems

[flybynight]

Hi everyone,

Here are three logarithm problems I am struggling with:

1. "Find the exact value of x satisfying the equation 3[sup:25wox7u4]x[/sup:25wox7u4] * 4 [sup:25wox7u4]2x+1[/sup:25wox7u4] = 6[sup:25wox7u4]x+2[/sup:25wox7u4]
Give your answer in the form ln a/ln b where a and b are real numbers"
For this one, I have attempted to divide the second term out of the first half and then taken the natural log of that, also I have attempted to consolidate the terms by attempting to get the three terms with the same base, but neither were successful.
2. "Solve 2(5[sup:25wox7u4]x+1[/sup:25wox7u4]) = 1 + (3/5[sup:25wox7u4]x[/sup:25wox7u4]), giving the answer in the form of a+log [sub:25wox7u4]5[/sub:25wox7u4] b"
For this one, I tried to substitute in 'x' for 5[sup:25wox7u4]x[/sup:25wox7u4], thereby getting 2(x[sup:25wox7u4]2[/sup:25wox7u4])-1 = 3/x and worked it out from there. I am not certain if I did something wrong, and thus did not get the right answer, or if the method is inherently flawed.
3. "Find an expression for the sum of the first 35 terms of the series ln x[sup:25wox7u4]2[/sup:25wox7u4] + ln x[sup:25wox7u4]2[/sup:25wox7u4]/y + ln x[sup:25wox7u4]2[/sup:25wox7u4]/y[sup:25wox7u4]2[/sup:25wox7u4] + ln x[sup:25wox7u4]2[/sup:25wox7u4]/y[sup:25wox7u4]3[/sup:25wox7u4] + ln x[sup:25wox7u4]2[/sup:25wox7u4]/y[sup:25wox7u4]4[/sup:25wox7u4]. Giving your answer in the form ln x[sup:25wox7u4]m[/sup:25wox7u4]/y[sup:25wox7u4]n[/sup:25wox7u4], where m,n are natural numbers."
For this one, I got ln x[sup:25wox7u4]2a[/sup:25wox7u4] - ln y[sup:25wox7u4]s[/sup:25wox7u4], a being the number of terms, s being the sum of the numbers between 1-34 (595). Thus, I get 2a ln x - s ln y. So I get 70 ln x - 595 ln y. From this I get ln x[sup:25wox7u4]70[/sup:25wox7u4] / y[sup:25wox7u4]595[/sup:25wox7u4]. In this problem, I am looking for confirmation of my answer.

Thank you very much,

Peter

 

[tkhunny]

Here are three logarithm problems I am struggling with:

1. "Find the exact value of x satisfying the equation 3[sup:2l4tj1il]x[/sup:2l4tj1il] * 4 [sup:2l4tj1il]2x+1[/sup:2l4tj1il] = 6[sup:2l4tj1il]x+2[/sup:2l4tj1il]
Give your answer in the form ln a/ln b where a and b are real numbers"
For this one, I have attempted to divide the second term out of the first half and then taken the natural log of that, also I have attempted to consolidate the terms by attempting to get the three terms with the same base, but neither were successful.

It appears that you have forgotten that these are labled "logarithm problems".

Try this:

x(log(3)) + (2x+1)log(4) = (x+2)log(6)

Now what?

 

[flybynight]

To be honest, I really have no idea what to do next. The problem is that I have to answer in ln a/ln b. I could solve it easily by graphing if it weren't for that.

Peter

 

[tkhunny]

Can you solve this one...

3x + 4(2x+1) = 6(x+2)

 

[Deleted member 4993]

or this one (solve for x):

A*x + (2x+1)*B = (x+2)*C............where A,B & C are some constants.

 

[soroban]

Hello, Peter!

I solved these problems at another site.
But for those you who just tuned in . . .

\(\displaystyle \text{1. Solve for }x\!:\;\;3^x\cdot4^{2x+1} \;=\;6^{x+2}\)

\(\displaystyle \text{Give your answer in the form: }\frac{\ln a}{\ln b}\text{, where }a\text{ and }b\text{ are real numbers.}\)

\(\displaystyle \text{We have: }\;3^x\cdot\left(2^2\right)^{2x+1} \;=\;(2\cdot3)^{x+2} \quad\Rightarrow\quad 3^x\cdot2^{4x+2} \;=\;2^{x+2}\cdot3^{x+2}\)

\(\displaystyle \text{Divide by }3^x\!\cdot\!2^{x+2}\!:\quad 2^{3x} \;=\;3^2\)

\(\displaystyle \text{Take logs: }\;\ln\left(2^{3x}\right) \;=\;\ln(9) \quad\Rightarrow\quad 3x\!\cdot\!\ln(2) \;=\;\ln(9)\)

\(\displaystyle \text{Therefore: }\;x \;=\;\frac{\ln(9)}{3\ln(2)} \;=\;\frac{\ln(9)}{\ln(2^3)} \;=\;\boxed{\frac{\ln(9)}{\ln(8)}}\)

\(\displaystyle \text{2. Solve: }\:2\cdot5^{x+1} \;= \;1 + \frac{3}{5^x}\)

\(\displaystyle \text{Give your answer in the form: }\:a + \log_5(b)\)

\(\displaystyle \text{Multiply by }5^x\!:\quad 2\cdot5^{2x+1} \;=\;5^x + 3\)

. . \(\displaystyle 2\cdot5\cdot5^{2x} - 5^x - 3 \;=\;0 \quad\Rightarrow\quad 10\cdot5^{2x} - 5^x - 3 \;=\;0\)

\(\displaystyle \text{Factor: }\;\left(2\cdot5^x + 1\right)\left(5\cdot5^x - 3\right) \;=\;0\)

\(\displaystyle \text{And we have: }\;\begin{array}{cccccccccccc}2\cdot5^x + 1 &=&0 & \Rightarrow & 5^x &=& \text{-}\frac{1}{2} & \text{No real roots} \\ \\[-4mm] 5\cdot5^x - 3 &=&0 & \Rightarrow & 5^{x+1} &=& 3 \end{array}\)

\(\displaystyle \text{Therefore: }\;x +1\;=\;\log_5(3) \quad\Rightarrow\quad x \;=\;\boxed{-1 + \log_5(3)}\)

\(\displaystyle \text{3. Find an expression for the sum of the first 35 terms of the series:}\)

. . \(\displaystyle \ln(x^2) +\ln\left(\frac{x^2}{y}\right) +\ln\left(\frac{x^2}{y^2}\right) + \ln\left(\frac{x^2}{y^3}\right) + \cdots\)

\(\displaystyle \text{Give your answer in the form }\ln\left(\frac{x^m}{y^n}\right)\text{, where }m,n\text{ are natural numbers}\)


For this one, I got: ln(x[sup:1244u1xc]2a[/sup:1244u1xc]) - ln(y[sup:1244u1xc]s[/sup:1244u1xc]),

a being the number of terms, s being the sum of the numbers between 1-34 (595).
Thus, I get: 2a ln(x) - s ln(y). So I get: 70 ln(x) - 595 ln(y).
From this I get: ln(x[sup:1244u1xc]70[/sup:1244u1xc] / y[sup:1244u1xc]595[/sup:1244u1xc]).

In this problem, I am looking for confirmation of my answer. . . . . Absolutely correct!

\(\displaystyle \text{We have: }\;\ln\left(\frac{x^2}{1}\cdot\frac{x^2}{y}\cdot\frac{x^2}{y^3}\cdot\frac{x^2}{y^4} \cdots \frac{x^2}{y^{34}}\right)\)

. . \(\displaystyle = \;\ln\left(\frac{x^{(2+2+2+\cdots+2)}}{y^{(1+2+3+\cdots+34)} }\right) \;= \;\ln\left(\frac{x^{35\cdot2}} {y^{(34\cdot35)/2}} \right) \;=\;\ln\left(\frac{x^{70}}{y^{595}}\right)\)

 

[tkhunny]

Again, we'll just never know if the student learned anything, will we?!

 

[soroban]

I can only hope that he learned some techniques for handling these problems.

If he didn't ... and he is simply copying and handing in homework ... that's his loss.
He will certain fail all his exams . . . and that is NOT my concern.

I am here to help people who want help . . . regardless of their motivation.
I do not have the time nor the inclination to interrogate each and every poster
what they intend to do with the information I give them.

I trust that 99% of them are not the slackers that some people fear are exploiting us.


I am truly sick of this sniping.
I've abandoned other math sites who have promoted this "We don't do homework" attitude,
as if our knowledge is too precious, too sacred to give out to just anybody who asks.
And the tutors openly scold people for failing to ask questions in the "proper" way.

Observation: OF COURSE, 99.9% of the questions are Homework Questions!
Why else would any of them have questions on mathemtics?
Intellectual curiosity? .A hobby? .Masochism? .Penance?

 

[stapel]

I am truly sick of this sniping.

And some tutors weary of expending great time and effort to craft a "teaching" reply designed to lead the student through the concepts and toward ongoing success (the "teach a man to fish" sort of reply), only to find that, while he was laboring, somebody else posted a quickie hand-in solution (the "give a man a fish" sort of reply). After all, some might posit that, if looking over one fully-worked solution were all the student needed to understand the material, then he'd have gained all he needed from reviewing his text and his lecture notes. Many feel that students possibly might learn a bit more if they eventually engage in some "doing", rather than sticking with only "receiving". Of course, these positions (and the research and documentation appearing to support them) may be entirely incorrect....

It might be noted that those who hang around to see the results of what is posted have all-too-often seen such very sad outcomes when the students are simply "given fish": The students may follow up by posting complete homework sets, becoming abusive when the rest of their homework isn't quickly done for them; or they may claim to "totally understand" the fully-worked exercise and then post another exercise of exactly the same sort, claiming to still have "no idea" how even to get started. Of course, these frequent results may be entirely unrepresentative and misleading....

Some tutors, having expressed the desire to see students grow in maturity, ability, and self-confidence, may be somewhat "sick" of hurtful outcomes. Of course, my supposition could be entirely wrong....

Eliz.

 

[tkhunny]

I am truly sick of this sniping.

You couldn't wait for one useful response from the student? Just one? I'm not asking for two.

There are now two outstanding questions that the student cannot be motivated to answer. You killed them. Do you really think that is appropriate, fair, or considerate? Now, if I'm badgering the student, which I sometimes do, I wouldn't have a problem with your new direction at all. Frankly, I would expect it.

I believe there is no evidence of me "sniping" except when you step on my outstanding questions. (There may be such evidence, but it wouldn't be normal for me to do that. I could have been having a bad day, or something.) I should think that the student's failure to respond to direct questions might be an indication that you, or anyone else, needn't waste any time on additional comment - an indication that you don't seem to appreciate.

My views. I welcome others'.