Question

8logt 81=32

Answer 1

Is the question 8 ( log(t)(81) )= 32 ?

Answer 2

These are the two log rules you'll need:
log (xy) = log(x) + log(y) and log x^y = y log(x)
8 (log(t) + log 81) = 32
8 (log(t) + log (3^4)) = 32
8log(t) + (8)(4)log(3) = 32
8log(t) + 32 log (3) = 32
8log(t) = 32 - 32 log(3) Now divide both sides by 8.
log(t) = 4 - 4 log (3) Now I used my calculator.
log(t) = 2.09151

10^(log t) = 10^2.09151
t = 123.46

Answer 3

Dear Studious Danielle,

Here's the question as I read it: 8logt 81=32; verbal instructions: Solve for the base t such that 8 (log to base t) of 81 = 32.

If that's the problem you wanted to solve, here's my approach:

Immediately divide both sides of the equation by 8.

Then (log to the base t) of 81 = 4

Since the two functions y = (base a)^x and y = (log to the base a) of x are inverse functions, we can write

t^(log to the base t) of 81 = t^4

This reduces to 81 = t^4.

t=?

Hint: 81=3^4,
so 3^4 = t^4.
Then base t must be what?