How do i write this as a single logarithim using the properties of logarithims? ${log}_{5} 9 + {log}_{5} 8 - 1$ $log_5 9+log_5 8-1$

Question

How do i write this as a single logarithim using the properties of logarithims? ${log}_{5} 9 + {log}_{5} 8 - 1$ $log_5 9+log_5 8-1$

1 Answer

Impact of this question

1107 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-10T21:04:45

${log}_{x} x \Leftrightarrow 1$ $log_x xhArr1$ and vice versa.

Explanation:

So we now have:
$= {log}_{5} 9 + {log}_{5} 8 - {log}_{5} 5$ $=log_5 9+log_5 8 -log_5 5$

We use the properties:
$log (a \times b) \Leftrightarrow log a + log b$ $log(axxb)hArrloga + logb$
$log (\frac{a}{b}) \Leftrightarrow log a - log b$ $log(a/b)hArrloga - logb$

So:
$= {log}_{5} (9 \times 8) - {log}_{5} 5$ $=log_5 (9xx8) -log_5 5$

$= {log}_{5} (\frac{9 \times 8}{5}) = {log}_{5} (\frac{72}{5}) = {log}_{5} 14.4$ $=log_5 ((9xx8)/5)=log_5 (72/5)=log_5 14.4$

Science

Math

Humanities

... and beyond

How do i write this as a single logarithim using the properties of logarithims? ${log}_{5} 9 + {log}_{5} 8 - 1$ $log_5 9+log_5 8-1$

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do i write this as a single logarithim using the properties of logarithims? log59+log58−1log_5 9+log_5 8-1

1 Answer

Explanation:

Related questions

Impact of this question

How do i write this as a single logarithim using the properties of logarithims? ${log}_{5} 9 + {log}_{5} 8 - 1$ $log_5 9+log_5 8-1$