Help! How do you integrate 2^logx? (1 Viewer)

nikitasuitsrevenge · Jan 1, 2014

I have tried to turn 2 into e^log2 but i keep getting it wrong. If someone can help, that would be great.

AAEldar · Jan 1, 2014

Let $u = log{x}$ and $du = \frac{1}{x} dx$ , then sub in and you'll have $x 2^{u}$ and re-arrange the initial substitution.

Carrotsticks · Jan 1, 2014

nikitasuitsrevenge · Jan 1, 2014

Thanks carrotsticks and AAelder

braintic · Jan 1, 2014

Carrotsticks said:
$\int 2^{\ln x} = \int 2^{\frac{\log_2 x}{\log_2 e}} = \int \left ( 2^{\log_2 x} \right )^\frac{1}{\log_2 e} = \int x^\frac{1}{\log_2 e} $ now proceed to integrate as usual, as this is in the form $ \int x^n dx$

Hmm, not the standard way to do this question.
Kind of makes the integration unnecessarily messy.

asianese · Jan 1, 2014

I think that method is a good alternative that avoids substitutions which can get messy itself.

By substitution:

$u = \ln x \Rightarrow e^u = x \\ du = \frac{dx}{x} \\ I = \int 2^u e^u \mathop{du} \\ = \int (2e)^u \mathop{du} \\ = \int e^{\ln ((2e)^u} \mathop{du} \\ = \int e^{(\ln(2e))u} \mathop{du}$

and go from there.

Drongoski · Jan 1, 2014

$2^{ln x} = e^{ln(2^{lnx})} = e^{ln x \cdot ln 2} = (e ^{ln x} )^{ln 2} = x^{ln 2}$

$\therefore \int = \int x^{ln 2} dx = \frac {x^{ln 2 + 1}}{ln 2 + 1} + C$

Wonder if this answer is right?

asianese · Jan 1, 2014

Ooo interesting. Cool little identity:

$a^{\log_b c} = c^{\log_b a}$

$a^{\log_b c} = \exp(\ln(a^{\log_b c})) \\ = \exp(\log_b(c) \cdot \ln (a)) \\ = \exp\left( \dfrac{\ln c}{\ln b} \cdot \ln a\right)\\ = \exp\left( \ln c \cdot \dfrac{\ln a}{\ln b}\right) \\ = \exp(\ln c \cdot \log_b (a))\\ = (\exp(\ln c))^{\log_b a}= c^{\log_b a}.$

Carrotsticks · Jan 1, 2014

braintic said:
Hmm, not the standard way to do this question.
Kind of makes the integration unnecessarily messy.

The reason why I used this method is because the question was posted in the Extension 1 section, so I assumed that a non-substitution method was wanted.

braintic · Jan 2, 2014

Carrotsticks said:
The reason why I used this method is because the question was posted in the Extension 1 section, so I assumed that a non-substitution method was wanted.

Guess what - I didn't read the question properly.

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Help! How do you integrate 2^logx? (1 Viewer)

nikitasuitsrevenge

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AAEldar

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Carrotsticks

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nikitasuitsrevenge

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braintic

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asianese

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asianese

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