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I'm having trouble seeing number 3 clearly. Or rather, I'm having trouble seeing a proof. I've tested a series of numbers and it works, but I'm having trouble seeing why. Can anyone clarify?

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Here is where I get confused:

I start with this log:

$$ \log_2 4^1 = 2$$

now if I raise both sides to the 2nd power then:

$$ \log_2 4^{1^2} = 2^2$$ $$ = \log_2 4^2 = 4$$ $$ = 2 \cdot \log_2 4 = 4$$

but if I raise both sides to the power of 1.5 then: $$ \log_2 4^{2 \cdot 1.5} \ne 4^{1.5} \ne 8$$

Jwan622
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2 Answers2

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By definition: $$ a^{\log_a x}=x $$

Now, if: $$ \log_a (x^y)=\alpha $$ we have $$ a^\alpha=x^y =\left( a^{\log_a x}\right)^y = a^{y\log_a x} $$ so $$ \alpha=y\log_a x $$

Emilio Novati
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Notice that using property $1$, we have $$\begin{align}\log_a(x^y)&=\log_a(\underbrace{x\cdot x\cdots\cdot x}_{y\,\text{times}})\\&=\underbrace{\log_a(x)+\log_a(x)+\cdots+\log_a(x)}_{y\,\text{times}}\\&=y\log_a(x)\end{align}$$

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