Solving 4-log(3-x)=3: Is -10/3 the Answer?

[wat2000]

4-log(3-x)=3 i got -10/3. is this correct? and if it isn't can someone show me the steps?

 

[Sethric]

That isn't what I am getting. What is the base for your logarithm? 10? Show us your steps so we can see what you are doing.

 

[SammyS]

4-log(3-x)=3 i got -10/3. is this correct? and if it isn't can someone show me the steps?

In order for this equation to be true, doesn't log(3-x) need to equal 1 ?

 

[wat2000]

Now I am getting x= -7. The way i got that answer is like this. first i subtracted 4 from both sides. this gave me -log(3-x)=-1. Then i divided both sides by -1. This gave me log(3-x)=1. I am not really sure what this next step is but its what they did in the book. It just looks like they took the parentheses away from the 3-x and divided by base 10 or something, anyway this step makes the problem look like this: 3-x=10^1. Then i subtract 3 from both sides this gives me -x=7 I divide by -1 and this gives me x = -7. is this right? If its not where am i going wrong?These are the same steps the my precalc book use.

 

[Mark44]

x = -7 is correct.
The step after log(3 - x) = 1 is not division. It uses the idea that the log function (base 10 in this case) is the inverse of the exponential function (10^x in this case).

In symbols, this is log(x) = y <==> x = 10^y

log(3 - x) = 1
<==> 3 - x = 10¹

or 3 - x = 10, so x = -7

 

[wat2000]

Thanks I appreciate the help.