Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(2^{13/5}))
i got log (8*14^{3/5}) / Log (7^{3/5 }*1/(2^{13/5 })) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(2^{13/5}))
i got log (8*14^{3/5}) / Log (7^{3/5 }*1/(2^{13/5 })) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
This is a simple exercise in properties of the exponential. Do you not know that $\displaystyle log(a^x)= x log(a)$ and $\displaystyle log(ab)= log(a)+ log(b)$?
[QUOTE=chewydrop;830170]Log(8)+3/5Log(14) - 3/5Log(7)+Log(1/(2^{13/5}))[/tex]
$\displaystyle 8= 2^3$ so [tex]log(8)= 3 log(2). $\displaystyle 14= 2(7)$ so $\displaystyle (3/5)log(14)= (3/5)log(2)+ (3/5)log(7)$. $\displaystyle 1/2^{13/5}= 2^{-13}{5}$ so $\displaystyle log(1/2^{13/5})= -(13/5)log(2)$
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*14^{3/5}) / Log (7^{3/5 }*1/(2^{13/5 })) = Log(2)
Just wondering if you can solve this without using a calculator since my exams forbid calculators.
Hello, chewydrop!
$\displaystyle \text{Simplify: }\:\log(8)+\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)+\log\left(\frac{1}{2^{\frac{13 }{5}}}\right)$
Re-group terms: .$\displaystyle \left[\log(8) + \log\left(2^{-\frac{13}{5}}\right)\right] + \left[\tfrac{3}{5}\log(14) - \tfrac{3}{5}\log(7)\right] $
. . . $\displaystyle =\;\left[\log(2^3) + \log(2^{-\frac{13}{5}})\right] + \tfrac{3}{5}\left[\log(14) - \log(7)\right] $
. . . $\displaystyle =\; \log\left(2^3\cdot2^{-\frac{13}{5}}\right) + \tfrac{3}{5}\log\left(\frac{14}{7}\right)$
. . . $\displaystyle =\;\log\left(2^{\frac{2}{5}}\right) + \tfrac{3}{5}\log(2)$
. . . $\displaystyle =\; \tfrac{2}{5}\log(2) + \tfrac{3}{5}\log(2)$
. . . $\displaystyle =\; \log(2)$