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How do you prove that

\(\large{b^{\log_b a}=a}\)?

An algebraic proof would be good.

Note by Bloons Qoth 2 years ago

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2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

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Let \(x=\log_{b} a \)

Then, \[ x=\log_{b} a \implies b^{x} = a \quad (\text{ by definition }) \\\implies b^{\log_{b} a} = a \quad (\text{ substituting } x) \]

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We can prove this algebraically if we want to. But why do that? Isn't this obvious?

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Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

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TopNewestLet \(x=\log_{b} a \)

Then, \[ x=\log_{b} a \implies b^{x} = a \quad (\text{ by definition }) \\\implies b^{\log_{b} a} = a \quad (\text{ substituting } x) \]

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We can prove this algebraically if we want to. But why do that? Isn't this obvious?

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