log function....

**frozenflames** · October 2nd 2009, 03:52 PM

how would i get x? the previous method does not work.....

**pickslides** · October 2nd 2009, 04:00 PM

You already have x!

**frozenflames** · October 2nd 2009, 04:02 PM

Originally Posted by pickslides

You already have x!

well i meant in its simplest form...

**pickslides** · October 2nd 2009, 04:07 PM

It doesn't get much more simple

$x = 2^{\log_23+\log_28}$

$x = 2^{\log_23+\log_22^3}$

$x = 2^{\log_23+3\log_22}$

$x = 2^{\log_23+3\times 1}$

$x = 2^{\log_23+3}$

$x = 2^3\times 2^{\log_23}$

$x = 8\times 2^{\log_23}$

**frozenflames** · October 2nd 2009, 04:11 PM

i'm supposed to be getting a whole number i think

**Matt Westwood** · October 2nd 2009, 04:13 PM

Well it's nearly there. What's $2^{\log_2 x}$ for any $x$ ?

**frozenflames** · October 2nd 2009, 04:16 PM

Originally Posted by Matt Westwood

Well it's nearly there. What's $2^{\log_2 x}$ for any $x$ ?

honestly i'm not too sure, my guess would be 1?

**Soroban** · October 2nd 2009, 04:17 PM

Hello, frozenflames!

Simplify: . $x \:=\:2^{\log_23 + \log_28}$

We have: . $x \;=\;2^{\log_23+\log_28} \;=\; 2^{\log_2(3\cdot8)} \;=\;2^{\log_2(24)} \;=\;24$

**Matt Westwood** · October 2nd 2009, 04:22 PM

Originally Posted by frozenflames

honestly i'm not too sure, my guess would be 1?

This is the most fundamental truth about exponentials and logarithms that you really need to understand: one is the inverse of the other. That is, logarithms "undo" exponents, and exponents "undo" logarithms.

$a = \log_b c \iff c = b^a$ for any $a$ , and any $b, c$ positive. That's how logarithms are defined.

So $2^{\log_2 x} = \log_2 (2^x) = x$ .

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