The value of x has to be determined given that `log_4(x - 5) = -1`

`log_4(x - 5) = -1`

If `log_a x = b` , `x = a^b`

` `=> `x - 5 = 4^-1`

=> `x = 1/4 + 5`

=> `x = 5.25`

**The required value of x is 5.25**

Log4(X-5) =-1

LogCD= a ----- ca= D

X-5=4-1

X-5= 0.25 Add 5 to the both side.

x-5+5=0.25+5

X=5.25

**Equation:** `log_(4)(x-5) = -1`

**To find:** value of x

**Answer**: we will use the following properties:

- `x^(-1)=1/x`
- `log_(a)(x)=y``=> x = a^y`

now from the given equation

`log_(4)(x-5) = -1`

`=> (x-5) = 4^-1`

`=> x-5 = 1/4`

`=> x = 5+(1/4)`

`=> x = 5 + 0.25`

`=> x = 5.25`

` `

These are two helpful links that may prove useful both presently and in the future (I'm assuming you are just beginning to cover logarithms in class):

http://www.enotes.com/topics/logarithm-of-a-power

http://www.enotes.com/topics/logarithmic-equations

To get rid of the log, we must do the opposite function which is an exponent. Make both sides an exponent with a base for 4. this gets you

x - 5 = 4^-1 = 1/4

Now, you can solve for x just like any other linear function:

x = .25 + 5

x = 5.25