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I was asked to find $x$ when: $$\log_{9}\left(\frac{1}{\sqrt3}\right) =x$$ Step two may resemble: $${3}^{2x}=\frac{1}{\sqrt3}$$

I was not allowed a calculator and was told that it was possible. I put it into my calculator and found out that $x$=-0.25 but how do you get that?

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    $\begingroup$ The key here is that the base for the logarithm is $9$. A logarithm asks you what power the base must be taken to in order to get the argument, here $1/\sqrt{3}$. Perhaps you should start by expressing both the base and the argument as powers of $3$. $\endgroup$
    – hardmath
    Nov 3, 2017 at 1:26
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    $\begingroup$ Hint: how would you express $1/\sqrt{3}$ as a power of $9$? $\endgroup$
    – Jason
    Nov 3, 2017 at 1:26
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    $\begingroup$ $3$ takes offense at being called weird. $\endgroup$ Nov 3, 2017 at 1:27
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    $\begingroup$ Note that $1/\sqrt 3=3^{-1/2}=9^{-1/4}$ $\endgroup$ Nov 3, 2017 at 1:29
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    $\begingroup$ When writing latex, you need to use curly braces to group things that go in an exponent: 3^2x gives $3^2x$, but 3^{2x} gives $3^{2x}$. $\endgroup$
    – user14972
    Nov 3, 2017 at 4:28

10 Answers 10

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You are asked to find $\displaystyle \log_{9}\big(\frac{1}{\sqrt3}\big)$, which means a number $x$ such that $$9^x = \frac{1}{\sqrt3}$$

At this point, it is natural to express everything as powers of $3$. (I admit that for this to come naturally takes some skill and experience: for example, you need to be able to see that $9 = 3^2$ and that $\sqrt{3} = 3^{1/2}$.) So:

$$3^{2x} = 3^{-1/2}$$

At this point it is a simple equation:

$$2x = -\frac12,$$

or $$x = -\frac14.$$

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    $\begingroup$ The OP was reeeally close to getting there; he just gave up at the $\dfrac1{\sqrt{3}}$. $\endgroup$ Nov 3, 2017 at 7:03
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Use the change of base logarithm formula \begin{eqnarray*} \log_9 \frac{1}{\sqrt{3}} = \frac{ \log_3 3^{ -\frac{1}{2}}} {\log_3 3^2} = \color{red}{-\frac{1}{4}}. \end{eqnarray*}

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Since $(\sqrt{3})^{4} = 9$ then $$\log_{9}\left(\frac{1}{\sqrt{3}}\right) = \frac{1}{4} \, \log_{9}\left(\frac{1}{9}\right) = - \frac{1}{4} \, \log_{9}(9) = - \frac{1}{4}.$$

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  • $\begingroup$ How would I know log9(1/√3) is equal to 1/4log9(1/9) without a calculator? $\endgroup$
    – John D
    Nov 3, 2017 at 1:36
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    $\begingroup$ @JohnD Express $1/\sqrt3$ as $3$ to an integral power, i.e. $3^{-1/2}$, then use log rules. Will Jagy’s answer covers this. $\endgroup$ Nov 3, 2017 at 2:03
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There are rules.

$$ \log_B C = \frac{\log_A C}{\log_A B} $$ Here you want $A = 3.$

Also $$ \log_A \frac{1}{T} = - \log_A T, $$ $$ \log_A W^P = P \log_A W $$ Here you would be using $P = 1/2$

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Hint:

$\log_9(\frac{1}{\sqrt{3}})=x$ is equivalent to $9^x=\frac{1}{\sqrt{3}}$

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Start with a related problem that you should know the answer to: $$\log_{9}\left(9\right) =1$$

When you square root a number, you halve it's logarithm $$\log_{9}\left(3\right) =0.5$$ $$\log_{9}\left({\sqrt3}\right) =0.25$$

When you take the reciprocal of a number, you reverse the sign of it's logarithm.

$$\log_{9}\left(\frac{1}{\sqrt3}\right) =-0.25$$

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Observe that $9^{-{\frac{1}{4}}} = \frac{1}{\sqrt{3}}$. Then, the expression can be rewritten as $$\log_9 {9^{-\frac{1}{4}}}$$ Recall that the logarithm just finds the power the base needs to be raised to in order to be equal to the logarithm's operand. Since we need to raise $9$ to the power of $-\frac{1}{4}$ to result in $9^{-\frac{1}{4}}$, the logarithm above is equal to $-\frac{1}{4}$, which is equal to $-0.25$.

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There are already some great answers to this question, but just in case you need the basics of logs...

One nice way to think about logs is that it asks for the how many of a particular factor is in a number.

$$log_5(125)=3$$

Because there are three factors of 5 in the number 125.
$$125=5 \cdot 5 \cdot 5=5^3$$

Similarly, $$log_3(81)=4$$

Because $$81=9 \cdot 9=3 \cdot 3 \cdot 3 \cdot 3=3^4$$ Finding simple logs is just a matter of counting a particular factor.


If you have the log of a fraction like $$log_3 \left(\frac{1}{81}\right)$$ Then we are dealing with a negative number of factors. $$log_3 \left(\frac{1}{81}\right)=-4$$ Because $$\frac{1}{81}=3^{-4}$$ Similarly $$log_2 \frac{1}{8}=-log_2(8)=-3$$
We have one more concept to look at and then we'll be ready. $$log_{25} (5)$$ We can think of 5 as one half of a factor of 25. $$log_{25} (5)=\frac{1}{2}$$ Similarly $$log_{25} (125)=\frac{3}{2}$$ Because there are 3 half factors of 25 in 125. We can also look at $$log_{81}(3)=\frac{1}{4}$$ Since $$81=3 \cdot 3 \cdot 3 \cdot 3$$ We need four factors of 3 to make 81, so 3 is 1 out of the four factor of 3 we need to make an 81. $$log_5{\sqrt 5}=\frac{1}{2}$$ Because $\sqrt 5$ is half a factor of 5. $$log_{25}{\sqrt 5}=\frac{1}{4}$$ Because it takes four factors of $\sqrt{5}$ to make 25. $$\sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5 \cdot \sqrt 5=25$$ So $\sqrt{5}$ is one fourth of a factor of 25.
So looking at your problem by building up, $$log_9(3)=\frac{1}{2}$$ $$log_9(\sqrt{3})=\frac{1}{4}$$ $$log_9\left(\frac{1}{\sqrt{3}}\right)=-\frac{1}{4}$$

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$\log_{9}\left(\frac{1}{\sqrt3}\right) = $

$\log_{9}(3^{-\frac 12})=$

$\log_{9}((\sqrt{9})^{-\frac 12})=$

$\log_{9}((9^{\frac 12})^{-\frac 12}) = $

$\log_{9}(9^{\frac 12*(-\frac 12}) =$

$\log_9(9^{-\frac 14}) =$

$-\frac 14$

..... or .....

$\log_9(\frac 1{\sqrt 3}) =x$ So

$9^x = \frac 1{\sqrt 3}= \frac 1{\sqrt{\sqrt 9}} = \frac 1{\sqrt[4]{9}} = 9^{-\frac 14}$

$x = -\frac 14$

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I think the key here is to remember what the logarithm actually means: $\log_a(b)$ is the answer to the question "which power of $a$ equals $b$?" In your case you need to answer the question "which power of nine equals $1/\sqrt{3}$?" The answer to this is not too difficult to find: $9^{-1} = 1/9$ and taking square roots twice gives the answer.

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