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Old November 17th, 2009, 05:08 PM
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Exclamation General Statement

Let Logax = c and Logbx = d

I need to find the general statement that expresses Logabx in the form of c and d
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Old November 17th, 2009, 06:36 PM
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Hello, Bzaher!

Quote:
Let: .\begin{array}{ccc}\log_ax \:=\:c \\ \log_bx \:=\:d \end{array}

Find the \log_{ab}x in terms of c and d.

\begin{array}{cccccc}\log_ax \:=\:c & \Rightarrow& a^c \:=\:x & \Rightarrow & a \:=\:x^{\frac{1}{c}} & [1] \\
\log_bx\:=\:d & \Rightarrow & b^d \:=\:x & \Rightarrow & b \:=\:x^{\frac{1}{d}} & [2] \end{array}


Multiply [1] and [2]: .ab \:=\:x^{\frac{1}{c}}\cdot x^{\frac{1}{d}} \;=\;x^{\frac{1}{c} + \frac{1}{d}} \quad\Rightarrow\quad ab \;=\;x^{\frac{c+d}{cd}}

Therefore: .ab^{\frac{cd}{c+d}} \;=\;x \quad\Rightarrow\quad \log_{ab}x \;=\;\frac{cd}{c+d}


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Old November 17th, 2009, 09:23 PM
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Thank you so much for the answer, but why is it x to the power of 1/c and 1/d
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