Evaluate the logarithm without using a calculator

1. Without using tables [or calculators], find the value of \[\begin{align*} 2 \lg 5 - \frac{1}{2} &\lg 9 + \lg 0.12.\\ [\lg x &= \log_{10} x] \end{align*}\]

Using the logarithm laws, we can rewrite the first two terms as

\[\begin{align*} 2 \lg 5 &= \lg 5^2 = \lg 25\\ - \frac{1}{2}\lg 9 &= \lg 9^{-\frac{1}{2}} = \lg \frac{1}{\sqrt{9}} = \lg \frac{1}{3}. \end{align*}\]

Now we can evaluate the original expression

\[\begin{align*} \lg 25+\lg \frac{1}{3}+\lg \frac{12}{10} &= \lg \left( \frac{25\times12}{3\times10}\right)\\ &=\lg \left( 10\right)\\ &=1. \end{align*}\]

1. If \(\lg 2 = 0.30103\) and \(\lg 3 = 0.47712\), find, without using tables [or calculators], the values of \(\lg 6\) and \(\lg 15\).

Following the logarithm laws we can rewrite the logarithm of six as \[\lg 6 = \lg 2\times3 = \left(\lg 2\right)+\left(\lg 3\right).\]

Inserting the given values for \(\lg 2\) and \(\lg 3\) we therefore obtain \[\lg 6 = 0.30103 + 0.47712 \approx 0.7782.\]

Since both given values have a precision of 5 decimal places, we can approximate the solution to 4 decimal places.

Fifteen can be expressed as \[15=\frac{10\times3}{2}.\]

Since we know the base-10 logarithm of 2, 3 and 10 we can write

\[\begin{align*} \lg 15 &= \lg \frac{10\times3}{2}\\ &= \left( \lg 10 \right) + \left( \lg 3 \right) - \left( \lg 2 \right)\\ &\approx 1 + 0.47712 - 0.30103\\ &= 1.1761. \end{align*}\]

Everything You Need in One Place Homework problems? Exam preparation? Trying to grasp a concept or just brushing up the basics? Our extensive help & practice library have got you covered.

Learn and Practice With Ease Our proven video lessons ease you through problems quickly, and you get tonnes of friendly practice on questions that trip students up on tests and finals.

Instant and Unlimited Help Our personalized learning platform enables you to instantly find the exact walkthrough to your specific type of question. Activate unlimited help now!

Get the most by viewing this topic in your current grade. Pick your course now.

Video transcript

Let's give ourselves a little bit more practice with logarithms. So just as a little bit of review, let's evaluate log base 2 of 8. What does this evaluate to? Well, it's asking us or it will evaluate to the power or the exponent that I have to raise our base to, that I have to raise 2 to, to get to 8. So 2 to the first power is 2. 2 to the second power is 4. 2 to the third power is 8. So this right over here is going to be equal to 3. Fair enough. We did examples like that in the last video. Let's do something a little bit more interesting. And I'll color-code it. What is log base 8 of 2? Now, this is interesting. I'll give you a few seconds to think about it. Well, we're asking ourselves, or this will evaluate to, the exponent that I have to raise 8 to to get to 2. So let's think about that in another way. So we could say 8 to some power-- and that exponent that I'm raising 8 to is essentially what this logarithm would evaluate to. 8 to some power is going to be equal to 2. Well, if 2 to the third power is 8, 8 to the one-third power is equal to 2. So x is equal to 1/3. 8 to the one-third power is equal to 2, or you could say the cube root of 8 is 2. So in this case, x is 1/3. This logarithm right over here will evaluate to 1/3. Fascinating. Let's mix it up a little bit more. Let's say we had the log base 2. Instead of 8, let's put a 1/8 right over here. So I'll give you a few seconds to think about that. Well, it's asking us, or this will evaluate to, the exponent that I have to raise 2 to to get to 1/8. So if we said that this is equal to x, we're essentially saying 2 to the x power is equal to 1/8. Well, we know 2 to the third power-- let me write this down. We already know that 2 to the third power is equal to 8. If we want to get to 1/8, which is the reciprocal of 8, we just have to raise 2 to the negative 3 power. 2 to the negative 3 power is 1 over 2 to the third power, which is the same thing as 1/8. So if we're asking ourselves, what exponent do we have to raise 2 to to get to 1/8? Well, we have to raise it to the negative 3 power. So x is equal to negative 3. This logarithm evaluates to negative 3. Now let's really, really mix it up. What would be the log base 8 of 1/2? What does this evaluate to? Let me clean this up so that we have some space to work with. So as always, we're saying, what power do I have to raise 8 to to get to 1/2? So let's think about that a little bit. We already know that 8 to the one-third power is equal to 2. If we want the reciprocal of 2 right over here, we have to just raise 8 to the negative one-third. So let me write that down. 8 to the negative one-third power is going to be equal to 1 over 8 to the one-third power. And we already know the cube root of 8, or 8 to the one-third power, is equal to 2. This is equal to 1/2. So the log base 8 of 1/2 is equal to-- well, what power do I have to raise 8 to to get to 1/2-- is negative 1/3. Equal to negative 1/3. I hope you enjoyed that as much as I did.