Indices Equations Questions

**Lites** · April 10th 2010, 09:22 AM

Hello there Math master, (I wonder if I could become one later...)

I have this several questions in which I could not find the answer..I'm stuck.

If $10^m = 2$ and $10^n = 3$ , find the value of 10^3m-n

Now I have tried randomly and find out that m = 1/5, but I do not know the proper way how to find m and n respectively. The book said that we need to find the common factor of 10 and 2, but so far I could not find any because the number seems very weird...

There's also this another question which is nearly the same:

Given that $3^x = 5$ and $3^y = 7$ , find the value of 3^2x-3y, giving your answer in fraction.

Thank you for your kind help.

**maddas** · April 10th 2010, 09:25 AM

$10^m = 2$ so cubing $(10^m)^3 = 10^{3m} = 2^3 = 8$ .

So $10^{3m-n}={10^{3m} \over 10^n}={8\over3}$ .

**Lites** · April 10th 2010, 09:56 AM

Originally Posted by maddas

$10^m = 2$ so cubing $(10^m)^3 = 10^{3m} = 2^3 = 8$ .

So $10^{3m-n}={10^{3m} \over 10^n}={8\over3}$ .

Thank you for your time answering my question but there are several things I do not comprehend. You're cubing the 10^m but why you're not cubing the 10^n too? Could you explain it a little bit step-by-step?

**maddas** · April 10th 2010, 10:06 AM

Because you were asked for $10^{3m-n}$ , and not $10^{3m-3n}$ . Perhaps it would be clearer if I did...

$10^{3m-n} = {10^{3m}\over 10^n} ={10^{m+m+m}\over10^n}= {10^m\cdot10^m\cdot10^m\over10^n} = {2\cdot2\cdot2\over3}=\frac83$

?

**Lites** · April 10th 2010, 10:26 AM

Originally Posted by maddas

Because you were asked for $10^{3m-n}$ , and not $10^{3m-3n}$ . Perhaps it would be clearer if I did...

$10^{3m-n} = {10^{3m}\over 10^n} ={10^{m+m+m}\over10^n}= {10^m\cdot10^m\cdot10^m\over10^n} = {2\cdot2\cdot2\over3}=\frac83$

?

Ok I got it, so would this be correct: (the second question I posted in the first post)

square 3^x = 5 so that we got 3^2x = 25
cube 3^y = 7 so that we got 3^3y = 343

ultimately we have 25/344?

**harish21** · April 10th 2010, 10:30 AM

Originally Posted by Lites

Ok I got it, so would this be correct: (the second question I posted in the first post)

square 3^x = 5 so that we got 3^2x = 25
cube 3^y = 7 so that we got 3^3y = 343

ultimately we have 25/344 may be a typo!

yes. $\frac{25}{343}$

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