If $47^x = 8$ and $376^y = 128$ , find $\frac{3}{x}-\frac{7}{y}$ What I know:
$x={\log_{47}8}$ and $y=\log_{376}128$
How do I do this without using a calculator?
 A: Solve for $x$ and $y$ first --- for instance, $x = \dfrac{\log 8}{\log 47}$. Then evaluate the expression.
A: $47^x$=$2^3$ and $376^y$=$2^7$ take logarithms with base 2 both sides and you get $\frac{3}{x}$=$log_247$ and $\frac{7}{y}$=$log_2376$ just subtract them to get $\frac{3}{x}$ - $\frac{7}{y}$=$log_2\frac{47}{376}$ which is equal to $-3$.
A: $$\frac{3}{x}-\frac{7}{y}=\log_2{47}-\log_2{376}=\log_2{\frac{1}{8}}=-3$$
A: As Umberto said you can use the log function which helps in your case: 
Here is a detailed solution: 
Take the log on both side:  
$\log(47)^x = log(8)$,  using the properties of log you can write it as:  
$x \log(47) = \log(8)$, divide both sides with $\log(47)$   
$x  = \frac{\log(8)}{\log(47)} $
Simiralry, solve for y: 
$y = \frac{\log(128)}{\log(376)}$ 
Then substitute in the original question:  
$3 * \frac{\log(47)}{\log(8)} - 7 * \frac{\log(376)}{\log(128)}$
A: We want to calculate $$\frac{3}{x}-\frac{7}{y}=3\frac{log_247}{log_28}-7\frac{log_2376}{log_2128}=3\frac{log_247}{3}-7\frac{log_2376}{7}=log_2(\frac{47}{376})=log_2(\frac{1}{8})=-3$$
A: No logs, no calculator.  :)
$$47^x=8=2^3\Longrightarrow 47=2^{3/x}\\
376^y=128=2^7\Longrightarrow 376=2^{7/y}$$
Dividing, 
$$\frac {2^{3/x}} {2^{7/y}}=\frac {47}{376}=\frac 18\\
2^{3/x-7/y}=2^{-3}\\
\frac 3x-\frac 7y=\color{red}{-3}$$
