1. ## 3 Logarithm Problems

Hi everyone,
Here are three logarithm problems I have to solve:
1. "Find the exact value of x satisfying the equation 3^x*4^(2x+1) = 6^(x+2)
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^[x+1]) = 1 + (3/5^x), giving the answer in the form of a+log base 5 b"
For this one, I tried to substitute in 'x' for 5^x, thereby getting 2(x^2)-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^2 + ln x^2/y + ln x^2/y^2 + ln x^2/y^3 + ln x^2/y^4. Giving your answer in the form ln x^m/y^n, where m, n are natural numbers."
For this one, I got ln x^2a - ln y^s, 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^70 / y^595. In this problem, I am looking for confirmation of my answer.
Thank you very much,
Peter

2. Please don't post in multiple locations.

You have a hint on the first one. Feel free to demonstrate its solution and we can move on to #2.

3. Hello, Peter!

Whew! . . . These were certainly challenging.

1. Find the exact value of $x$ for: $3^x\cdot4^{2x+1} \:= \:6^{x+2}$

Give your answer in the form $\frac{\ln a}{\ln b}$ where $a,b$ are real numbers.
Write it in terms of 2's and 3's.

We have: . $3^x\cdot(2^2)^{2x+1} \;=\;(2\cdot3)^{x+2}$

. . . . . . . . . $3^x\cdot 2^{4x+2} \;=\;2^{x+2}\cdot3^{x+2}$

Divide by $2^{x+2}\!\cdot\!3^x\!:\;\;\;2^{3x} \;=\;3^2 \quad\Rightarrow\quad 2^{3x} \;=\;9$

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

Therefore: . $x \;=\;\frac{\ln(9)}{3\!\cdot\!\ln(2)} \;=\; \frac{\ln(9)}{\ln(2^3)} \;=\;\boxed{\frac{\ln(9)}{\ln(8)}}$

2. Solve: . $2\cdot5^{x+1} \;= \;1 + \frac{3}{5^x}$

Give the answer in the form: . $a+\log_5(b)$

Multiply by $5^x\!:\;\;2\cdot5^{2x+1} \;=\;5^x + 3 \quad\Rightarrow\quad 2\cdot5\cdot5^{2x} - 5^x - 3 \;=\;0$

We have: . $10\cdot5^{2x} - 5^x - 3 \;=\;0$

Factor: . $\left(2\!\cdot\!5^x + 1\right)\left(5^x - 3\right) \;=\;0$

Then: . $\begin{Bmatrix}2\!\cdot\!5^x + 1 &=&0 & \Rightarrow & 5^x &=& \text{-}\frac{1}{2} & \leftarrow&\text{no real roots} \\ \\[-4mm]
5^x - 3 &=&0 & \Rightarrow & 5^x &=& 3 & \Rightarrow & \boxed{x \;=\;\log_5(3)}\end{Bmatrix}$

3. Find an expression for the sum of the first 35 terms of the series:

. . $\ln(x^2) + \ln\left(\frac{x^2}{y}\right) + \ln\left(\frac{x^2}{y^2}\right) + \ln\left(\frac{x^2}{y^3}\right) + \ln\left(\frac{x^2}{y^4}\right) + \hdots$

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

We have: . $\ln\left(\frac{x^2}{1}\right) + \ln\left(\frac{x^2}{y}\right) + \ln\left(\frac{x^2}{y^2}\right) + \ln\left(\frac{x^2}{y^3}\right) + \cdots + \ln\left(\frac{x^2}{y^{n-1}}\right)$

. . $= \;\ln\left(\frac{x^2}{1}\cdot\frac{x^2}{y}\cdot\fr ac{x^2}{y^2}\cdot\frac{x^2}{y^4} \cdots \frac{x^2}{y^{n-1}}\right)$ . $= \;\ln\left(\frac{x^{(2+2+2+\cdots+2)}} {y^{(1 +2+3+\cdots +[n-1])}} \right)$ . $= \;\ln\left(\frac{x^{2n}}{y^{n(n-1)/2}}\right)$

If $n=35$, we have: . $\boxed{\ln\left(\frac{x^{70}}{y^{595}}\right)}$

4. Thank you Soroban!

5. I am delighted that you managed to get someone else to do your homework for you. Too bad we'll never know if you learned anything. Go answer my questions on the other site. Let's see if you have a clue.