Ask your own question, for FREE!
Mathematics 10 Online
OpenStudy (anonymous):

help with 2 math problems... 1. Solve the exponential equation 125^(8x-2)=150 A. -0.1203 B. 0.3797 C.0.3705 D.2.1297 2. Solve the equation. log(x+5) - logx =3 A. 0.0505 B. 0.0050 C. 2.5 D.199.8

OpenStudy (whpalmer4):

For the first one, a helpful thing to remember is that \[\log n^a = a \log n\]Use that to take the log of both sides.

OpenStudy (whpalmer4):

For the second, remember that \[\log a - \log b = \log(\frac{a}{b})\]

OpenStudy (anonymous):

I'm really bad at logs as a whole. But i will try

OpenStudy (whpalmer4):

don't worry, I've given you good hints :-)

OpenStudy (anonymous):

so for the first one would i divide 125 from both sides??

OpenStudy (whpalmer4):

\[125^{8x-2} = 150\]You can't usefully divide out the 125. Doing so would just give you \[125^{8x-3} = \frac{150}{125}\]

OpenStudy (whpalmer4):

You want to take the log of both sides, which would give you \[(8x-2) \log 125 = \log 150\]Does that look like something you could solve for \(x\)?

OpenStudy (anonymous):

how would i solve for x

OpenStudy (whpalmer4):

let's pretend that those logs aren't there. we'll replace log 125 with C and log 150 with D: \[(8x-2)C = D\]Can you solve that for \(x\)?

OpenStudy (whpalmer4):

It's just \[(8x-2) = \frac{D}{C}\]\[8x=2+\frac{D}C\]\[x=\frac{1}{8}(2+\frac{D}{C})\]right?

OpenStudy (anonymous):

right

OpenStudy (whpalmer4):

Okay, well, that's exactly how we would do it with the logs there! \[(8x-2)\log 125 = \log 150\]\[(8x-2) = \frac{\log 150}{\log 125}\]\[8x=2+\frac{\log 150}{\log 125}\]\[x = \frac{1}{8}(2+\frac{\log 150}{\log 125})\] and now it's just an exercise for your calculator

OpenStudy (anonymous):

i dont have a calculator. Except for the one on my phone. what i got im pretty sure its wrong

OpenStudy (whpalmer4):

actually, you don't need a calculator :-) log 150 = the number that you raise 10 to to get 150. 10^2 = 100. 10^3 = 1000. so the log of 150 is going to be a bit more than 2. Similarly, log 125 = the number you raise 10 to to get 125. Obviously, that's going to be somewhere between 2 and whatever you got for log 150. So if we divide log 150 by log 125, that's going to be something like 2.2/2.1, maybe — final result is pretty darn close to 1. 1+2 = 3. 3/8 = 0.375. Any answers close to that?

OpenStudy (anonymous):

B or C?

OpenStudy (whpalmer4):

Ah, okay, we have two answers that are in the ballpark. But our answer is going to be > 0.375 because log 150 / log 125 is > 1 so we'll have 2 + 1.something = 3.something / 8. That means C is the correct choice. If you have a web browser, google will answer as a calculator. go to google.com and type "log 150" into the search field and hit return.

OpenStudy (anonymous):

2.17609125906 thats what i got from google

OpenStudy (whpalmer4):

you can type the whole expression into the search field: (2+log 150/log 125)/8

OpenStudy (anonymous):

0.37972011469 so is it B?

OpenStudy (whpalmer4):

yes, that's correct!

OpenStudy (whpalmer4):

Now on to the second problem :-)

OpenStudy (whpalmer4):

\[\log(x+5)-\log(x) = 3\]By my hint, we can rewrite the left side as the log of a fraction: \[\frac{\log(x+5)}{\log(x)} = 3\]

OpenStudy (whpalmer4):

Uh, sorry, I formatted that incorrectly! \[\log{(\frac{x+5}{x})} = 3\]

OpenStudy (whpalmer4):

Now we can undo the log by raising the base of the log (10) to each side: \[10^{\log(\frac{x+5}{x})} = 10^3\]

OpenStudy (whpalmer4):

but raising the base of the log (10) to the log just cancels out, giving us \[\frac{x+5}{x} = 10^3 \]

OpenStudy (anonymous):

okay undo the log,, what happens next?

OpenStudy (whpalmer4):

now it's just straight algebra and arithmetic, no logs to be seen :-)

OpenStudy (anonymous):

thanks, i believe ill be able to figure it out now. I have to log off :( thanks for your help tho

OpenStudy (whpalmer4):

you bet. good luck with logs in the future!

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!