# how do you solve log 5 (0+1)-3?

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You need to evaluate the following logarithm `log_5 (0+1) - 3` , such that:

`log_5 (0+1) = log_5 1`

Since `log_5 1 = 0` yields:

`log_5 (0+1) - 3 = 0 - 3 = -3`

You may also evaluate the expression, such that:

`log_5 (0+1) - 3 = log_5 1 - 3*1`

Replace `log_5 5` for 1, such that:

`log_5 (0+1) - 3 = log_5 1 - 3*log_5 5`

Use the power property of logarithms, such that:

`log_5 (0+1) - 3 = log_5 1 - log_5 5^3`

Converting the difference of logarithms into the logarithm of quotient yields:

`log_5 (0+1) - 3 = log_5 1/(5^3)`

`log_5 (0+1) - 3 = log_5 (5^(-3)) `

Taking out the power of argument yields:

`log_5 (0+1) - 3 = -3log_5 5 = -3*1 = -3`

**Hence, evaluating the given expression containing logarithms, yields **`log_5 (0+1) - 3 = -3.`

*Best answer as selected by question asker.*

We'll re-write the equation:

3^(x-5) - 2 = 5^0

Let's recall that a number raised to the 0 power has the result 1.

3^(x-5) - 2 = 1

We'll add 2 both sides:

3^(x-5) = 2+1

3^(x-5) = 3

Because the bases are matching, we'll use the one to one property:

x-5 = 1

We'll add 5 both sides:

x = 5+1

**x = 6**

We'll verify the solution into equation:

3^(6-5) - 2 = 5^0

3^1 - 2 = 1

3-2 = 1

1 = 1