How would I write
1-2log(7)x as a single logarithm
Choices:
log(7) x^2/7
log(7) 7x^2
log(7) 7/x^2
log (7) x^2
Thank you
1-2log(7)x
= log77 - log7x^2
= log7(7/x^2)
Thank you
To write 1-2log(7)x as a single logarithm, you need to use the properties of logarithms. Specifically, you can use the property of logarithmic addition, which states that log(a) + log(b) is equal to log(ab).
In this case, we have 1 - 2log(7)x. Let's break it down step by step:
1 - 2log(7)x = log(7)1 - log(7)(x^2)
(The logarithm base remains the same throughout the simplification process. In this case, it is log base 7.)
Now, log(7)1 can be simplified further:
log(7)1 = 0
(A logarithm of 1 to any base is always equal to 0.)
Substituting, we get:
1 - 2log(7)x = 0 - log(7)(x^2)
= -log(7)(x^2)
Therefore, the answer is -log(7)(x^2). None of the given choices match this answer, so it seems that there might be a mistake in the answer choices provided.