# Thread: Assignment Help Needed Exponents and Logarithms

1. ## Assignment Help Needed Exponents and Logarithms

If anyone could help me with any of these questions, that would be awesome.

7. Express the function y = 5^x as an equivalent function with base 3.

8. Suppose money can be invested at 3.75% compunded semi-annually. Determine how long it takes for a principal to triple in value.

9. The decibel (dB) is used to measure the loudness of a sound. The equation D = 10logI models the decibel level of a sound whose intensity is I watts per square metre (W/m^2). The decibel levels of a subway train and normal conversation are 115 dB and 60 dB, respectively. How many times as intense as normal conversation is the noise of a subway train?

10. (a) solve the equation (2^x)^2 - 4(2^x) + 3 = 0
(b) Graph the function y = 2^2x - 4(2^x) + 3
(c) How is the graph in part (b) related to your answer in part (a)?
(d) Describe the graph in part (b). Explain why it has shape.

11. Prove logab = (logcb) / (logca). Hint: let x = logab, y = logcb, and z = logca.

2. Originally Posted by philipsach1
7. Express the function y = 5^x as an equivalent function with base 3..
From the definition of $\displaystyle \log_b(A)$ we have:

$\displaystyle 5=3^{\log_3(5)}$,

so:

$\displaystyle y=3^{x \log_3(5)}$

RonL

3. Originally Posted by philipsach1
.

9. The decibel (dB) is used to measure the loudness of a sound. The equation D = 10logI models the decibel level of a sound whose intensity is I watts per square metre (W/m^2). The decibel levels of a subway train and normal conversation are 115 dB and 60 dB, respectively. How many times as intense as normal conversation is the noise of a subway train?
Already answered in this thread, also look at the other solutions there as a number of the questions asked there are the same as yours.

RonL

4. Hello, Philip!

Here's #11 . . .

11. Prove: .$\displaystyle \log_ab \:=\:\frac{\log_cb}{\log_ca}$ . . We don't need their hint.

Let: .$\displaystyle \log_ab \:=\:x$ . Then: .$\displaystyle a^x\:=\:b$

Take logs of both sides (base $\displaystyle c$): .$\displaystyle \log_c(a^x) \:=\:\log_cb$

Then we have: .$\displaystyle x\cdot\log_ca \:=\:\log_cb\quad\Rightarrow\quad x \:=\:\frac{\log_cb}{\log_ca}$

. . Therefore: .$\displaystyle \log_ab \:=\:\frac{\log_cb}{\log_ca}$

5. Hello again, Philip!

Here's #8 . . .

8. Suppose money can be invested at 3.75% compunded semi-annually.
Determine how long it takes for a principal to triple in value.

Your expected to know the Compound Interest formula: .$\displaystyle A \;=\;P(1 + i)^n$

where: .$\displaystyle \begin{Bmatrix}P & = & \text{principal invested} \\ i & = & \text{periodic interest rate} \\ n & = & \text{number of periods} \\ A & = & \text{final amount} \end{Bmatrix}$

The interest rate is 3.75% compounded semi-annually.
. . Hence: .$\displaystyle i \,=\,\frac{0.0375}{2}\,=\,0.01875$

Semi-annually for $\displaystyle k$ years means $\displaystyle 2k$ periods.

The principal is $\displaystyle P$ dollars which will grow to $\displaystyle 3P$ dollars.

Substitute into the formula: .$\displaystyle 3P \;=\;P(1.01875)^{2k}\quad\Rightarrow\quad(1.01875) ^{2k} \:=\:3$

Take logs: .$\displaystyle \ln(1.01875)^{2k} \:=\:\ln(3)\quad\Rightarrow\quad2k\cdot\ln(1.01875 )\:=\:\ln(3)$

Then: .$\displaystyle k \;=\;\frac{\ln(3)}{2\ln(1.01875)} \;=\;29.57013043$

Therefore, the investment will take about $\displaystyle 30$ years to triple in value.

6. Originally Posted by philipsach1
If anyone could help me with ...

10. (a) solve the equation (2^x)^2 - 4(2^x) + 3 = 0
(b) Graph the function y = 2^2x - 4(2^x) + 3
(c) How is the graph in part (b) related to your answer in part (a)?
(d) Describe the graph in part (b). Explain why it has shape.
...
Hello,

to solve this equation use substitution: $\displaystyle t=2^x$. Then your equation changed to:

$\displaystyle t^2-4t+3=0 \Longleftrightarrow t = 1 \ \vee \ t=3$

Re-substitute:

$\displaystyle 1=2^x \Longrightarrow x = 0$ or
$\displaystyle 3=2^x \Longrightarrow x = \log_2(3) =\frac{\ln(3)}{\ln(2)} \approx 1.585$

to (b)I've attached a diagram of the graph

to (c): You have calculated the zeros of the function.

to (d): As you easily can see, the graph has a horizontal asymptote y = 3 if x approaches the neative infinity. Explanation:

$\displaystyle \lim_{x \rightarrow -\infty}y=\lim_{x \rightarrow -\infty}{(2^{2x})}-4 \cdot \lim_{x \rightarrow -\infty}{(2^x)}+3 = 0 - 0 +3 =3$

EB