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?
$\endgroup$ 24 Answers
$\begingroup$$\log_{49}{\sqrt{7}}=\dfrac{1}{4}$ means that $49^{\frac{1}{4}}={\sqrt{7}}$
In other words, when you take the fourth root of $49$, you should get $\sqrt{7}$, not $7$.
Note that $\Large49^{\frac{1}{4}}=(7^2)^{\frac{1}{4}}=7^\frac{2}{4}=7^{\frac{1}{2}}=\sqrt{7}$
$\endgroup$ 4 $\begingroup$No, you get $\sqrt 7$ as you should.
$$\log_{49}\sqrt 7 =\log_{49}7^{\frac12}=\frac12\log_{49}7=\frac12\cdot\frac12=\frac14$$
$\endgroup$ 2 $\begingroup$Solve: $49^x = \sqrt{7} \Rightarrow 49^{2x} = 7 \Rightarrow \frac{\log7}{\log 49} = 2x \Rightarrow \frac{1}{2}=2x \Rightarrow x = \frac{1}{4}$
$\endgroup$ $\begingroup$To get used to logarithm rules, try to relate them to exponent rules.
For example, think of a logarithm as the answer to the question "a to what power equals b". So the 2 following statements are equivalent to each other.
$\log a^?=b \iff ?=\log_ab$
Then try to convert logarithms to exponents, and manipulate them with the power rules that you're already familiar with.
$$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.
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