# Logarithms: What are the steps to evaluate $\log_5 3125$?

I am having a hard time understanding how to solve logarithmic functions.

For example $\log_5 3125$

This how I would solve it:

1) $5^x = 3125$

2) Answer is $5$. However, I can not use a calculator on my test thus how can I solve this function in a more formal way?

Also. how would I solve $\log_{36} \sqrt 6$?

• ... because several small powers $a^b$, including $5^5$, should be known by heart? Commented Jun 28, 2014 at 16:33

1) Just go through the multiplications of $5$ until you reach $3125$:

5^1 = 5

5^2 = 25

5^3 = 125

5^4 = 625

5^5 = 3125

You can do multiplication without a calculator...

2)

Note that $\sqrt{6}=6^{1/2}$ and apply the rule $\log x^y=y\log x$ then repeat as in 1.

• How do you know that $3125$ is perfect integer power of $5$? Commented Jun 28, 2014 at 16:31
• 1/2 log 36^6=x , Now what? Commented Jun 28, 2014 at 16:36
• Solve for the log and then multiply 1/2 by 1/2? Commented Jun 28, 2014 at 16:42
• Sometimes I've been amazed to see what undergraduate students cannot do without a calculator. I've seen a student grab a calculator when he needed to find $15\times 8$. And others like that. Commented Jun 28, 2014 at 16:50
• @labbhattacharjee, I know that 5 divides 3125 because if it didn't then it would be awfully cruel to have such a question on a non-calculator paper. Commented Jun 28, 2014 at 17:21

$$3125=5\cdot625=5\cdot(25)^2=5\cdot(5^2)^2=5\cdot5^4=5^5$$

and using $\displaystyle\log_ab=\frac{\log_ca}{\log_cb}$ and $\displaystyle\log_b(d^y)=y\log_bd$ when each logarithm remains defined

$$\log_{36}\sqrt6=\frac{\log_6(6^{\frac12})}{\log_6(6^2)}=\frac{\frac12\cdot1}{2\cdot1}=\cdots$$

The first step is to find the prime factorization $3125$. Then you find out whether or not it's a perfect power of $5$. You don't have to recognize it, as long as you can divide it by $5$ repeatedly.

The same trick would work if you wanted to know $\log_3 2187$. Just start dividing by $3$, and see what happens:

$2187=3\cdot 729=3\cdot 3\cdot 243 = 3\cdot 3\cdot 3\cdot 81$

At some point, you recognize a power of $3$, such as, $81=3^4$. Now you know that $2187=3^7$.

For the second one, think this way: We see a $36$ and a $\sqrt{6}$. Both of these can be expressed as powers of $6$, so we do that. Then just proceed as in @lab_bhattacharjee's answer, i.e., use the change-of-base formula with $6$ as the common base.

Intrestingly, You would require to us calculator for log when you are using Logarithms in problem solving like calculating area,volume etc. Simple straight forward questions like these require the use of fundamental logarithmic rules

Log basically evaluates the exponent(e) wrt to a base (b).

For example, $log_{10} 1000=3$ base being 10. That is, 3 powers to 10 and you will reach 1000.

So Generally, $log_b (value)= e$ such that, $b^e=value$

Rule 1 : $log_b A + log_b B= log_a AB$

Rule 2: $log_b A - log_b B= log_a \frac{A}{B}$ and vice versa.

Rule 3: Base conversion

$log_bc$=$\frac{log_ab}{log_ac}$

It is always good to convert base.

Coming back to your second question,

Let, $log_{36}\sqrt6 =y$

$\implies 36^y=\sqrt6$

$\implies 6^{2y}=6^{\frac{1}{2}}$

When, base are same both sides, exponents can be equated to one other

So, $2y=\frac{1}{2}$

Or, $y=\frac{1}{4}$

For a non calculator exam some values of log_{10} values by heart, i.e. the prime values of 2,3,5,7,11,13,17,19... Being an undergraduate myself I can tell you that no question involving any other factors was asked if calculator/log table was not provided. (even 11,13,17,19 are barely used)

Than you need to know basic log formulae. The one you use here is that log (a x b) = log a + log b (Rule 1 in SId's post)

So as long as you can factorize the number you can easily calculate log without any calculator.

Example: Log of 1001 in base 10. 1001 = 7x 11x 13

log 7 = 0.845, log 11 = 1.041, log 13 = 1.113, log 1001 = 2.999,

From a calculator I find the value as log(1001) = 3.00043408. So it is good enough.

just do what people did before calculators were invented.

$$5^x = 3125$$ $$5^x = 1\cdot5\cdot5\cdot5\cdot5\cdot5$$ $$5^x=5^5$$ $$x=5$$

As for $$\log_{36}\sqrt{6}$$, try to express everything in terms of powers of $$36$$. $$\log_{36}\sqrt{6} = \log_{36}\sqrt{\sqrt{36}}=\log_{36}\sqrt{36^{1/2}}=\log_{36}(36^{1/2})^{1/2}=\log_{36}36^{\frac{1}{2}\cdot \frac{1}{2}}=\log_{36}36^{\frac{1}{4}}=\frac{1}{4}$$