If $47^x = 8$ and $376^y = 128$ , find $\frac{3}{x}-\frac{7}{y}$

Question

What I know: $x={\log_{47}8}$ and $y=\log_{376}128$
How do I do this without using a calculator?

Answer 1

Solve for $x$ and $y$ first --- for instance, $x = \dfrac{\log 8}{\log 47}$. Then evaluate the expression.

Answer 2

$47^x$=$2^3$ 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 $-3$.

Answer 3

$$\frac{3}{x}-\frac{7}{y}=\log_2{47}-\log_2{376}=\log_2{\frac{1}{8}}=-3$$

Answer 4

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:
$3 * \frac{\log(47)}{\log(8)} - 7 * \frac{\log(376)}{\log(128)}$

Answer 5

We want to calculate $$\frac{3}{x}-\frac{7}{y}=3\frac{log_247}{log_28}-7\frac{log_2376}{log_2128}=3\frac{log_247}{3}-7\frac{log_2376}{7}=log_2(\frac{47}{376})=log_2(\frac{1}{8})=-3$$

Answer 6

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, $$\frac {2^{3/x}} {2^{7/y}}=\frac {47}{376}=\frac 18\\ 2^{3/x-7/y}=2^{-3}\\ \frac 3x-\frac 7y=\color{red}{-3}$$