1. ## Fixing log errors.

Trying to fix the log questions I got wrong. I've already fixed some, but I can't grasp these two in any way.

1) "log(15)1"

I did:

log(15)1 = x
15^x = 1
x = 1 ------> I corrected this with 15^0 = 1 ---> x = 0. But I can't grasp/remember the rule of anything to the zero becomes one. Does zero replace x?

2) "10^3x = 55"

I did:

10^3x = 55^1
3x = 55
x = 18.333 ---> I attempted to correct this and came up with x = log(10)55/3. I still resulted with x = 18.333 and I don't understand why this was still marked wrong.

2. Originally Posted by Mulya66
Trying to fix the log questions I got wrong. I've already fixed some, but I can't grasp these two in any way.

1) "log(15)1"

I did:

log(15)1 = x
15^x = 1
x = 1 ------> I corrected this with 15^0 = 1 ---> x = 0. But I can't grasp/remember the rule of anything to the zero becomes one. Does zero replace x?
log15(1) = x
1 = 15^x

Since we can say that 15^0 = 1, we can rewrite the problem as:
15^0 = 15^x

(Note that 15^0 replaced 1 in the problem 1 = 15^x, becoming 15^0 = 15^x.)

Therefore, x = 0.

2) "10^3x = 55"

I did:

10^3x = 55^1
3x = 55
x = 18.333 ---> I attempted to correct this and came up with x = log(10)55/3. I still resulted with x = 18.333 and I don't understand why this was still marked wrong.

Is this 10^(3x) = 55 or 10^(3) * x = 55? The notation you used was a bit unclear. But I'm going to assume you mean 10^(3x) = 55.

From this, you CANNOT say that 3x = 55. You can't set the exponent of the left hand side equal to the base of the right hand side (doing so is invalid and makes no sense). However, we can solve for x by taking the log of both sides:

log(10^(3x)) = log(55)
3x*log(10) = log(55)
3x = log(55)
x = log(55)/3

This will be an irrational answer that you'll have to plug into your calculator to find.