Question Video: Computing Logarithmic Expressions Using Laws of Logarithms | Nagwa Question Video: Computing Logarithmic Expressions Using Laws of Logarithms | Nagwa

Question Video: Computing Logarithmic Expressions Using Laws of Logarithms Mathematics

Calculate log₂ 192 − log₂ 3.

03:25

Video Transcript

Calculate log base two of 192 minus log base two of three.

This is an expression that we could just enter into our calculators. But actually, there’s a very elegant solution. In general, the difference of the logarithms of two quantities to the same base is equal to the logarithm to that base of the quotient of the two quantities.

The logarithms in our problem are to the same base. They’re both logarithms base two. And so their difference is the logarithm base two of their quotient, so log base two of 192 over three.

This general formula is one of the so called laws of logarithms. And we’ll come to prove it at the end of the video. 192 divided by three is 64. And so we have log base two of 64.

What is log base two of 64? If we call this 𝑥, then we can write this equation, which is in a logarithmic form, in exponential form as two to the power of 𝑥 equals 64. So really when I’m asking for log base two of 64, I’m asking two to the power of what equals 64.

And hopefully we recognise 64 as a power of two. In fact, it’s the sixth power of two. Two to the power of six is 64. We see that 𝑥, which is log base two of 64 remember, is six. There we have it, no calculator required. We just use one of the laws of logarithms, which we’ll now prove.

This is the law that we want to prove written in terms of the arbitrary values 𝑎, 𝑏, and 𝑐. All the logarithms we have have base 𝑎. And so a relatively natural thing to do is to find 𝑎 to the power of both sides.

So we now have 𝑎 to the power of this difference of logarithms is equal to 𝑎 to the power of the logarithm of the quotient. On the left-hand side, we have 𝑎 to the power of something minus something else. And we know from exponent laws that 𝑎 to the power of something minus something else is equal to 𝑎 to the power of that something divided by 𝑎 to the power of that something else.

So with something equal to log base 𝑎 of 𝑏 and something else equal to a log base 𝑎 of 𝑐, we get that 𝑎 to the power of log base 𝑎 of 𝑏 over 𝑎 to the power of log base 𝑎 of 𝑐 is equal to 𝑎 to the power of log base 𝑎 of 𝑏 over 𝑐. The right-hand side stays the same.

We now have three things of the form 𝑎 to the power of log base 𝑎 of 𝑥 in our equation. What is log base 𝑎 of 𝑥? Well if you remember from earlier, it’s the value that you have to raise 𝑎 to the power of to get 𝑥. This is another rule that is really helpful to remember. 𝑎 to the power of log base 𝑎 of 𝑥 is equal to 𝑥.

Applying this rule to our equation 𝑎 to the power of log base 𝑎 of 𝑏 is 𝑏. 𝑎 to the power of log base 𝑎 of 𝑐 is 𝑐. And 𝑎 to the power of log base 𝑎 of 𝑏 over 𝑐 is 𝑏 over 𝑐.

We have an equation which is clearly true. And if we follow the chain of logic in the other direction, we get the law of logarithms that we used to solve our problem.

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