Question

Si log a00 = m , hallar:
$log_{a}(a0)$
​

Answer 1

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RESPUESTA

$\Large{\boxed{ \boxed{ \log _{\text{a}} \overline{ \text{a0}} = \frac{ \text{m}-1}{\text{m} -2}}}}$

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EXPLICACIÓN

$\text{Si}: \: \: \log \overline{ \text{a00 }} \text{ = m}$

$\implies \: \: \: \log {( \text{a} \cdot 100)} \text{ = m}$

Se cumple que:

$\boxed{ \log \text{(a}\cdot \text{b)=} \log \text{a} + \log \text{b} }$

Entonces:

$\implies \: \: \: \log {( \text{a} \cdot 100)} \text{ = m}$

$\implies \: \: \: \log \text{a} + \log 100 \text{ = m}$

$\implies \: \: \: \log \text{a} + 2 \text{ = m}$

$\implies \: \: \: \log \text{a} \text{ = m}-2$

Se cumple:

$\boxed{ \log_{ \text{b}} \text{a} = \frac{ 1}{ \log_{ \text{a}} \text{b}} }$

Entonces:

$\implies \: \: \: \log \text{a} \text{ = m}-2$

$\implies \: \: \: \frac{1}{\log _{\text{a}} 10} \text{ = m}-2$

$\implies \: \: \: \log _{\text{a}} 10 = \frac{1}{\text{m}-2}$

Hallar:

$\implies \: \: \: \log _{\text{a}} \overline{ \text{a0}}$

$\implies \: \: \: \log _{\text{a}}\text{(a} \cdot \text{10)}$

$\implies \: \: \: \log _{\text{a}}\text{a} + \log _{\text{a}}\text{10}$

$\implies{1 + \log _{\text{a}}\text{10}}$

$\implies \: \: \: 1 + \frac{1}{ \text{m} - 2}$

$\implies \: \: \: \frac{ \text{m} - 2}{ \text{m} - 2} + \frac{1}{ \text{m} - 2}$

$\implies \: \: \: \frac{ \text{m} - 2+1}{ \text{m} - 2 }$

$\implies \: \: \: \frac{ \text{m} - 1}{ \text{m} - 2}$