### Video Transcript

Determine the value of the log of seven over 100 to the base two minus log of 56 over 23 to the base two plus log of 25 over 23 to the base two without using a calculator.

In order to answer this question, we firstly need to recall two of our laws of logarithms. If two logs have the same base, in this case, π, then log of π₯ plus log of π¦ is equal to log of π₯π¦. And log of π₯ minus log π¦ is equal to log of π₯ divided by π¦. Letβs consider the first part of our expression: log of seven over 100 minus log of 56 over 23.

As both of these have the same base, we can use the second law. We can divide seven over 100 by 56 over 23. Dividing two fractions is the same as multiplying by the reciprocal of the second fraction. We can, therefore, rewrite this as seven over 100 multiplied by 23 over 56. Seven and 56 are both divisible by seven as 56 divided by seven is equal to eight. In order to multiply two fractions, we multiply the numerators and then multiply the denominators. This means that log of seven over 100 minus log of 56 over 23 simplifies to log of 23 over 800.

We are now left with log of 23 over 800 plus log of 25 over 23. Once again, as both of these terms have a base of two, we can use one of the laws of logarithms, in this case, the first law. We need to multiply 23 over 800 by 25 over 23. There is a 23 on the top and the bottom. So these can cancel. 25 and 800 are both divisible by 25. 800 divided by 25 is equal to 32. Our expression simplifies to log of one over 32 to the base two.

We now need to consider how we can work out the value of this logarithm without using a calculator. Logarithms are linked to exponents in the following way. If π¦ is equal to log π₯ to the base π, then π₯ is equal to π to the power of π¦. In this question, our base π is equal to two. And the value of π₯ is one over 32. This means that one over 32 is equal to two to the power of π¦. Two to the power of five is equal to 32. As negative powers or exponents give us the reciprocal of the ordinary power, then two to the power of negative five is equal to one over 32. Our value for π¦ is, therefore, negative five. We can, therefore, conclude that log of one over 32 to the base two is equal to negative five.

This means that the value of our initial expression log of seven over 100 minus log of 56 over 23 plus log of 25 over 23 is equal to negative five.