How to evaluate logs using the general log rule

Solving logs always follows the same rules

We already know how to evaluate simple logs like $\log_2{8}$, because we understand that this log is asking us the question “To what exponent do we have to raise $2$ in order to get $8$, or

$2^x=8$

And since all the values in a problem like this one are whole numbers, it’s pretty easy to see that $x=3$. But log problems can get a little more complicated than this, and that’s what we want to talk about here.

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How to solve logs when the base is greater than the argument

In all our examples so far, the argument has been greater than the base. In $\log_2{8}$, the base is $2$ and the argument is $8$, so $8>2$ and argument > base. But what happens when the base is greater than the argument?

$\log_{27}{3}$

When we convert that into an exponent, we get

$27^x=3$

To solve this equation, we want to make the bases on each side equivalent to one another. We know that $27$ is the same as $3^3$, so we’ll rewrite the equation.

$(3^3)^x=3$

$3^{3x}=3$

$3^{3x}=3^1$

If the bases are equal, then the exponents must also be equal in order for the equation to be true.

$3x=1$

$x=\frac13$

So we can say that

$\log_{27}{3}=\frac13$

How to evaluate logs using the general log rule

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Solving logs by converting them to an exponential function

Example

Find the value given by the log.

$\log_{243}{3}$

We’ll convert to the alternative form.

$243^x=3$

We know that $243$ is the same as $3^5$, so

$(3^5)^x=3$

$3^{5x}=3$

$3^{5x}=3^1$

Since the bases are equivalent, the only way to make this equation true is for the exponents to also be equivalent.

$5x=1$

$x=\frac15$

So we can say that

$\log_{243}{3}=\frac15$

When the bases are equivalent, the only way to make The equation true is for the exponents to also be equivalent.

How to simplify the log when the argument is a fraction

Sometimes the argument will be a fraction, like this:

$\log_2{\frac{1}{64}}$

As always, we still always want to change this into the equivalent equation.

$2^x=\frac{1}{64}$

From here, we’ll manipulate the $1/64$.

$2^x=\frac{1}{2^6}$

$2^x=2^{-6}$

The bases are equivalent, so the exponents must be equivalent. Therefore, $x=-6$ and

$\log_2{\frac{1}{64}}=-6$

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Let’s try another example.

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Example

Find the value given by the log.

$\log_5{\frac{1}{625}}$

Write the expression in the alternative form.

$5^x=\frac{1}{625}$

From here, we’ll manipulate the $1/625$.

$5^x=\frac{1}{5^4}$

$5^x=5^{-4}$

The bases are equivalent, so the exponents must be equivalent. Therefore, $x=-4$ and

$\log_5{\frac{1}{625}}=-4$

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This method for solving logs will always work. If we can get the bases equal to one another, then we can also set the exponents equal to each other. Let’s show a summary of the steps with one last example, in which we solve $\log_{32}{16}$.

$\log_{32}{16}$

$32^x=16$

$(2^5)^x=2^4$

$2^{5x}=2^4$

$5x=4$

$x=\frac45$

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