# Thread: Can you help me solve these so I can prepare for a test?

1. ## Can you help me solve these so I can prepare for a test?

Most of these I can't do so I really need the answer to see how you did it in reverse. If you can't solve all of them I'd appreciate anything you can do to help.

1. If $y=\left ( \frac{1}{2} \right )^{2}$ is reflected in the y-axis state what would its equation be:

$y=-\left ( \frac{1}{2} \right )^{2}$?

2. Evaluate the following:

$log_{4}500-3log_{4}5$

no idea

3. What is the inverse of: $y=2^{x-1}$

$y=-2^{-x+1}$?

4. Solve:

a) $2^{x-2}+2^{x} =320$
no idea

b) $log_{2}\left ( 3x+1 \right )=4$

$3x+1=2^{4}$

$\frac{3x}{3}=\frac{15}{3}$

$x=5$?

c) $log_{2}(3x+2)+log_{3}(x-1)=3$

once again have no idea

5. Convert the following: $y=log_{2}(8x-16)^{2}$ into the form $y=alog_{3}(x-p)+q$ (if this makes no sense i might have typed it wrong)

EDIT: One more for now:

Solve: and express the answer correct to three decimal places.

I can attempt this one but im not sure how to calculate it

2. ## Re: Can you help me solve these so I can prepare for a test?

Too many questions for a single thread. In the future no more than three.

Originally Posted by Rayzala
1. If $y=\left ( \frac{1}{2} \right )^{2}$ is reflected in the y-axis state what would its equation be:

$y=-\left ( \frac{1}{2} \right )^{2}$?
y = 1/4 is not a typical function to graph. Is there an x missing in here somewhere?

Originally Posted by Rayzala
2. Evaluate the following:

$log_{4}500-3log_{4}5$

no idea
500 = 5 x 10^2. So log(500) = ?

Originally Posted by Rayzala
3. What is the inverse of: $y=2^{x-1}$

$y=-2^{-x+1}$?
To get an inverse you switch x and y. So if you have $\displaystyle y = 2^{x - 1}$ then you needs to solve $\displaystyle x = 2^{y - 1}$ for y. How do you get the y - 1 out of the exponent?

Try that out for now.

-Dan

3. ## Re: Can you help me solve these so I can prepare for a test?

Hello, Rayzala!

$\displaystyle \text{4(a) Solve: }\:2^{x-2} + 2^x \:=\:320$

Factor: .$\displaystyle 2^{x-2}(1 + 2^2) \:=\:320 \quad\Rightarrow\quad 5\cdot2^{x-2} \:=\:320 \quad\Rightarrow\quad 2^{x-2} \:=\:64$

. . . . . . $\displaystyle 2^{x-2} \:=\:2^6 \quad\Rightarrow\quad x-2 \:=\:6 \quad\Rightarrow\quad x \:=\:8$

$\displaystyle \text{4(b) Solve: }\:\log_2(3x+1) \:=\:4$

$\displaystyle 3x+1 \:=\:2^4 \quad\Rightarrow\quad 3x+1 \:=\:16 \quad\Rightarrow\quad 3x \:=\:15 \quad\Rightarrow\quad x \:=\:5$

$\displaystyle \text{5. Convert the following: }\: y\:=\:\log_2(8x-16)^2$

$\displaystyle y \;=\;\log_2\big[8(x-2)\big]^2 \;=\;\log_2\big[2^3(x-2)\big]^2 \;=\;\log_2\big[(2^3)^2(x-2)^2\big]$

$\displaystyle y \;=\;\log_2\big[(2^6)(x-2)^2\big] \;=\;\log_2(2^6) + \log_2(x-2)^2 \;=\;6 + 2\log_2(x-2)$

$\displaystyle \text{6. Solve to 3 decimal places: }\:3^x \:=\:300$

Take logs: .$\displaystyle \ln(3^x) \:=\:\ln(300) \quad\Rightarrow\quad x\ln(3) \:=\:\ln(300)$

. . . . . . . . . $\displaystyle x \;=\;\frac{\ln(300)}{\ln(3)} \;=\;5.191806549 \;\approx\;5.192$

4. ## Re: Can you help me solve these so I can prepare for a test?

Originally Posted by topsquark
Too many questions for a single thread. In the future no more than three.

Sorry Dan you were right, put a 2 instead of an x.. it should read: Would it be: ? You completely lost me with the 2nd one though.. not sure how to continue from 'log(500) ='

as for the third one... to get y-1 out of the exponent wouldn't I square both sides?But I don't know how to get to the final answer. (Edit: turns out i read the question wrong and i was just supposed to graph the inverse which is much easier.. dont really need the equation but if you can help me i'd like to see how anyway)

I will make sure to make another thread for the other questions because you said there's too many here. Thanks to the both of you.

5. ## Re: Can you help me solve these so I can prepare for a test?

Originally Posted by Rayzala
Sorry Dan you were right, put a 2 instead of an x.. it should read: Would it be: ? You completely lost me with the 2nd one though.. not sure how to continue from 'log(500) ='

as for the third one... to get y-1 out of the exponent wouldn't I square both sides?But I don't know how to get to the final answer.

I will make sure to make another thread for the other questions because you said there's too many here. Thanks to the both of you.
Absolutely not! The inverse of an exponential function is a logarithmic function. First, I would simplify by writing the function as \displaystyle \begin{align*} y = 2^{-x} \end{align*}. Now swap the x and y values and solve for y.

6. ## Re: Can you help me solve these so I can prepare for a test?

Originally Posted by Rayzala
To reflect a set of points over the y-axis means that your are changing x to -x. So you are correct!

Originally Posted by Rayzala
[ATTACH=CONFIG]
$\displaystyle log_4(500) = log_4(5 \cdot 10^2) = log_4(5) + log_4(10^2) = log_4(5) + 2 \cdot log_4(10)$

Can you finish from here?

-Dan