# Thread: Is it possible to not use a calculator and solve this problem?

1. ## Is it possible to not use a calculator and solve this problem?

Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))

i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)

Just wondering if you can solve this without using a calculator since my exams forbid calculators.

2. ## Re: Is it possible to not use a calculator and solve this problem?

Originally Posted by chewydrop
Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))

i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)

Just wondering if you can solve this without using a calculator since my exams forbid calculators.
You mean you *did* use a calculator? Where??

3. ## Re: Is it possible to not use a calculator and solve this problem?

Originally Posted by Matt Westwood
You mean you *did* use a calculator? Where??
(8*143/5) / (73/5 *1/(213/5 )) into the calculator and got 2.0000001

4. ## Re: Is it possible to not use a calculator and solve this problem?

This is a simple exercise in properties of the exponential. Do you not know that $log(a^x)= x log(a)$ and $log(ab)= log(a)+ log(b)$?

[QUOTE=chewydrop;830170]Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(213/5))[/tex]
$8= 2^3$ so [tex]log(8)= 3 log(2). $14= 2(7)$ so $(3/5)log(14)= (3/5)log(2)+ (3/5)log(7)$. $1/2^{13/5}= 2^{-13}{5}$ so $log(1/2^{13/5})= -(13/5)log(2)$

So you have 3 log(2)+ (3/5)log(2)+ (3/5)log(7)- (3/5)log(7)- 13/5 log(2)

Now, it's just arithmetic.

i got log (8*143/5) / Log (73/5 *1/(213/5 )) = Log(2)

Just wondering if you can solve this without using a calculator since my exams forbid calculators.

5. ## Re: Is it possible to not use a calculator and solve this problem?

Hello, chewydrop!

$\text{Simplify: }\:\log(8)+\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)+\log\left(\frac{1}{2^{\frac{13 }{5}}}\right)$

Re-group terms: . $\left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]$

. . . $=\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]$

. . . $=\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)$

. . . $=\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)$

. . . $=\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)$

. . . $=\; \log(2)$

6. ## Re: Is it possible to not use a calculator and solve this problem?

Originally Posted by Soroban
Hello, chewydrop!

Re-group terms: . $\left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right]$

. . . $=\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right]$

. . . $=\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)$

. . . $=\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)$

. . . $=\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)$

. . . $=\; \log(2)$
For some reason when i regrouped it on my own i got a different expression:
Log(8)+Log(2-13/5)-3/5Log(7)+3/5Log(14)

would this still work? because im stuck already