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I cannot understand why $\log_{49}(\sqrt{ 7})= \frac{1}{4}$. If I take the $4$th root of $49$, I don't get $7$.

What I am not comprehending?

Sujaan Kunalan
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Frustrated with Math
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4 Answers4

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$\log_{49}{\sqrt{7}}=\dfrac{1}{4}$ means that $49^{\frac{1}{4}}={\sqrt{7}}$

In other words, when you take the fourth root of $49$, you should get $\sqrt{7}$, not $7$.

Note that $\Large49^{\frac{1}{4}}=(7^2)^{\frac{1}{4}}=7^\frac{2}{4}=7^{\frac{1}{2}}=\sqrt{7}$

Sujaan Kunalan
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No, you get $\sqrt 7$ as you should.

$$\log_{49}\sqrt 7 =\log_{49}7^{\frac12}=\frac12\log_{49}7=\frac12\cdot\frac12=\frac14$$

MPW
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  • Why did you move the 1/2 to the left of the operation log? Moreover, where did the second 1/2 come from? And how were you able to multiply them to get 1/4. I am utilizing "Pre-calculus by Stewart". – Frustrated with Math Mar 31 '14 at 01:22
  • If I have $\log_ab^c$, I can rewrite it as $c\log_ab$, so that's how the first $\frac{1}{2}$ came from. The second one came from $\log_{49}7=\frac{1}{2}$. (which came from $49^{\frac{1}{2}}=7$) – Sujaan Kunalan Mar 31 '14 at 01:23
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Solve: $49^x = \sqrt{7} \Rightarrow 49^{2x} = 7 \Rightarrow \frac{\log7}{\log 49} = 2x \Rightarrow \frac{1}{2}=2x \Rightarrow x = \frac{1}{4}$

Mr.Fry
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To get used to logarithm rules, try to relate them to exponent rules.

For example, think of a logarithm as the answer to the question "a to what power equals b". So the 2 following statements are equivalent to each other.

$\log a^?=b \iff ?=\log_ab$

Then try to convert logarithms to exponents, and manipulate them with the power rules that you're already familiar with.

$$x=\log_ab$$ $$a^x=b$$ $$(a^x)^c=b^c$$ $$a^{cx}=b^c$$ $$cx=\log_ab^c$$ $$c\log_ab=\log_ab^c$$

Try doing this with the other exponent rules and logarithms will be a snap.

John Joy
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