Question Video: Evaluating Logarithms | Nagwa Question Video: Evaluating Logarithms | Nagwa

# Question Video: Evaluating Logarithms Mathematics • Second Year of Secondary School

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What is the value of logβ (1/128)?

02:11

### Video Transcript

What is the value of the logarithm to the base two of one over 128?

To find the value of the logarithm to base two of one over 128, we first note that one over 128 is equal to 128 to the power of negative one. So, this means that the logarithm to the base two of one over 128 is the logarithm to the base two of 128 to the power negative one. And now, we can apply the power rule for logarithms. This says that the log to the base π of π raised to the power π₯ is equal to π₯ times log to the base π of π. And what this means is that if our argument has an exponent π₯, we can bring this number down and multiply our logarithm by it.

In our case, our exponent is negative one so that we have negative one times log to the base two of 128. Now, recalling our definition of a logarithm, if the logarithm to the base π of π is equal to π, then π raised to the power π is equal to π. And what this means is that π is the number of times we multiply the base π by itself to give π. So now, weβre looking for the negative of the number of times we multiply the base two by itself to get 128. So, we need to find a value of π for which two raised to the πth power is 128.

And if we look at our powers of two, we can see that two to the seventh power is 128. So that our π is equal to seven. This means that log to the base two of 128 is equal to seven. Therefore, negative log to the base two of 128 is negative seven. And we have our solution log to the base two of one over 128 is negative seven.

Itβs worth noting, perhaps, that we could have done this in a slightly different way, using the fact that two to the power of seven is 128. Writing log to the base two of one over 128 is equal to negative log to the base two of 128, we can write this as negative log to the base two of two raised to the power of seven. And using our power rule for logarithms once more, we get negative seven log to the base two of two. We know from our laws of logarithms that log to the base π of π is equal to one. And we have here log to the base two of two, so thatβs equal to one. So, again, we reach our answer of a negative seven.

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