Question

Prove that `(1 + Log_5 8 + Log_5 2)/log_5 6400 = 0.5` Please note the base number of each log is 5 and not 10. I continuously get `1/80`, can someone please assist?

Answer 1

`1/2`

Explanation:

`6400 = 25*256 = 5^2*2^8`
`=> log(6400) = log(5^2) + log(2^8) = 2 + 8 log(2)`
`log(8) = log(2^3) = 3 log(2)`
`=> (1+log(8)+log(2))/log(6400) = (1+4 log(2))/(2+8log(2)) = 1/2`

Answer 2

Apply common logarithmic identities.

Explanation:

Let's start by rewriting the equation so it's easier to read:

Prove that:
`(1 + log_5 8 + log_5 2) / (log_5 6400) = 0.5`

First, we know that `log_x a + log_x b = log_x ab`. We use that to simplify our equation:

`(1 + log_5 8 + log_5 2) / (log_5 6400) = (1 + log_5 (8*2)) / (log_5 6400) = (1 + log_5 16) / (log_5 6400)`

That "`1+`" is getting in the way, so let's get rid of it. We know that `log_x x = 1`, so we substitute:

`(1 + log_5 16) / (log_5 6400) = (log_5 5 + log_5 16) / (log_5 6400)`

Using the same addition rule from before, we get:
`(log_5 5 + log_5 16) / (log_5 6400) = (log_5 5 * 16) / (log_5 6400) = (log_5 80) / (log_5 6400)`

Finally, we know that `log_x a = log_b a / log_b x`. This is commonly called the "change of base formula" - an easy to way to remember where the `x` and `a` go is that `x` is below the `a` in the original equation (because it's written smaller under `log`).

We use this rule to simplify our equation:

`(log_5 80) / (log_5 6400) = log_6400 80`

We can re-write the logarithm into an exponent to make it easier:

`log_6400 80 = x`
`6400^x = 80`

And now we see that `x = 0.5`, since `sqrt(6400) = 6400^0.5 = 80`.
`square`

You probably made the mistake that `(log_5 80) / (log_5 6400) = 80/6400 = 1/80`. Be careful, this is not true.