Need Help

[Tram]

Hi, this is my first time in this forum, and I really need help badly.
I scanned out the questions that I need help on.

I am stuck on these problems, so I hope there is someone who can give me step by step explanation on how to do these questions.

Thx.

 

[ChaoticLlama]

First of all, you really should post in the correct forum.

Secondly, please don't just give us a 'laundry list' of problems. We love to assist people in mathematics, we do not like to give out answers without the poster showing some work first.

For number 1) you have 2^(3x + 5) = 8^(887)
Make both of the bases equal, then use properties of exponents to solve hte problem

P.S. You should be able to do the calculator approximations or else you may be in trouble.

 

[Tram]

Sorry about putting this topic at the wrong section. I don't see any place for Pre. Calculus.

As for my questions, I looked at my notes and the notes showed me solutions
for problems that are easlier than this problem.

As for question one, my teacher told me not to use a calculator to do it, so I don't know any method to do it.

Thanks for tell me that I must try to do some work on it, so I will now do some work.

Edited: Umm...I want to make sure, as for question number 5, do I use the laws od logs: Log(AB) = Log(A) + Log(B) ?

 

[soroban]

Hello, Tram!

Welcome aboard!
Here are a few of them . . .

$\displaystyle 5)\;\log_5(\sqrt{125})\,+\,32^{-\frac{1}{5}}$

We have: $\displaystyle \,\log_5(125)^{\frac{1}{2}}\,+\,(2^5)^{-\frac{1}{5}}\;=\;\frac{1}{2}\cdot\log_5(125)\,+\,2^{(5)(-\frac{1}{5})}\;=\;\frac{1}{2}\cdot\log_5(5^3)\,+\,2^{-1}$

$\displaystyle \;\;\;=\;3\cdot\frac{1}{2}\cdot\log_5(5)\,+\,\frac{1}{2^1}\;=\;\frac{3}{2}\cdot1\,+\,\frac{1}{2}\;=\;2$

$\displaystyle 8)\;\tan\left(\frac{7\pi}{4}\right)\,+\,\cos\left(\frac{7\pi}{6}\right)$

You're expected to know the trig values for $\displaystyle 30^o,\;60^o,\;45^o$ and their variations.

$\displaystyle \tan\left(\frac{7\pi}{4}\right)\,+\,\cos\left(\frac{7\pi}{6}\right)\;=\;-1\,+\,\left(-\frac{\sqrt{3}}{2}\right)\;= \;-\frac{2\,+\,\sqrt{3}}{2}$

$\displaystyle 11)\;\cos\left(\frac{11\pi}{12}\right)$

Formula: $\displaystyle \,\cos(A\,+\,B)\;=\;\cos(A)\cdot\cos(B)\,-\,\sin(A)\cdot\sin(B)$

Since $\displaystyle \,\frac{11\pi}{12}\:=\:\frac{3\pi}{4}\,+\,\frac{\pi}{6}$

we have: $\displaystyle \,\cos\left(\frac{3\pi}{4}\,+\,\frac{\pi}{6}\right)\;=\;\cos\left(\frac{3\pi}{4}\right)\cdot\cos\left(\frac{\pi}{6}\right)\,-\,\sin\left(\frac{3\pi}{4}\right)\cdot\sin\left(\frac{\pi}{6}\right)$

$\displaystyle \;\;\;\;= \;\left(-\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)\,-\,\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right) \;= \;-\frac{\sqrt{6}}{4}\,-\,\frac{\sqrt{2}}{4}\;=\;-\frac{\sqrt{6}\,+\,\sqrt{2}}{4}$

 

[Tram]

Thank you soroban.
I found sloved number 11, and made a mistake before, and saw your answer which I when back to check and found the problem I did. Thanks again.

For number 5, I think I never learn to do that kind of problem, and as for number 8 I only have on fault, and it is converting radian to degree, if I can master on converting radian to degree, then I can do that problem easliy.

Thanks again, I will try to do the other rest on my own.

 

[stapel]

Thread moved to "Geometry". Questions are as follows:

. . . . .1) Solve: 2^{3x + 5} = 8⁸⁸⁷

. . . . .5) Find log₅(sqrt[125]) + 32^-1/5

. . . . .7) Find arccos(-sqrt[3]/2) + arctan(sqrt[3]/3)

. . . . .8) Evaluate tan(7pi/4) + cos(7pi/6)

. . . . .11) Use a sum or difference formula to find the exact
. . . . . . . .value of cos(11pi/12).

. . . . .22) Use your calculator to find a six-digit approximation
. . . . . . . .to the solution of the equation 3^x = 1707

. . . . .23) Use your calculator to find a six-digit approximation
. . . . . . . .to cos^-1(7/9) in radians.

 

[Tram]

I want to know one thing about (7pi/4).
I want to see if I am converting it to degree correctly, I use 180/4 = 45. Then 45 times 7, then I end up with 315. Then I use 360 minus 315, then I get 45, and use (sqrt 2)/2.

Is that correct?