How do you evaluate ${log}_{9} 729$ $log_9 729$ ?

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Answer 1 · 2016-08-07T13:35:09

When I write ${log}_{a} b = c$ $log_(a)b=c$ . I state explicitly that $a^{c} = b$ $a^c=b$

Typically we use sensible bases for logarithms such as $10$ $10$ or $e$ $e$ .

But with your problem we have ${log}_{9} 729$ $log_(9)729$ .

And thus if ${log}_{9} 729 = x$ $log_(9)729=x$ , then $9^{x} = 729 = 9^{3}$ $9^x=729=9^3$ .

Clearly ${log}_{9} 729$ $log_(9)729$ $=$ $=$ $3$ $3$ .

Answer 2 · 2016-08-07T18:42:48

$3$ $3$

In log form, the question being asked is :

"To what power must 9 be raised to equal 729?"

It really is a huge advantage to know all the powers up to 1,000 by heart.

In this case you will recognize $729 = 9^{3}$ $729 = 9^3$

Therefore ${log}_{9} 729 = 3$ $log_9 729 = 3$

Without knowing this, you will have to do the whole log route...

${log}_{9} 729 = x$ $log_9 729 = x$
$9^{x} = 729$ $9^x = 729$

#xlog9 = log729+

$x = \frac{log 729}{log 9} = 3$ $x = (log729)/(log9) = 3$

It really is easier to just learn them!

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