Logarithm help need explanation

**dzomberg** · January 7th 2013, 08:17 AM

Write 1 - 2log₇x as a single logarithm.
Thanks!

**Plato** · January 7th 2013, 08:27 AM

Originally Posted by dzomberg

Write 1 - 2log₇x as a single logarithm.
Thanks!

$\log_7(7)-\log_7(x^2)=\log_7\left(\frac{7}{x^2}\right)~.$

**shervlad** · January 7th 2013, 08:28 AM

$1{\rm{ }} - {\rm{ }}2{\log _7}x = {\log _7}7 - {\log _7}{x^2} = {\log _7}{7 \over {{x^2}}}$

**russo** · January 7th 2013, 08:30 AM

Hi, this kind of problems involves the use of properties. So you want that expression to be simplified in one logarithm.

First, let's write everything as a logarithm. We know that: $log_n (n) = 1$ so we could say that $n = 7$ (in our case). Thus $1 = log_7 (7)$

Then, we have to put, for the second part, "inside" the logarithm. So we also know that: $a \cdot log_n(x) = log_n(x^a)$ . Finally $2 \cdot log_7(x) = log_7(x^2)$

We put it altogether (using the substraction of logarithm property of the same base) $log_n(x)-log_n(y) = log_n\left(\frac{x}{y}\right)$ :

$1 - 2\cdot log_7(x) = log_7(7) - log_7(x^2) = log_7\left(\frac{7}{x^2}\right)$

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