#1**0 **

log (e^(2x+2))=7

Using one of the rules of logs - x^(logx(y))= y

You can rewrite the log as - e^(log e^(2x+2)) = e^7 -- to isolate whats in the bracket (2x+2)

You'll now be left with - 2x+2=e^7

Solving for X 2x=e^7-2

x=(e^7-2)/2

Because e is an actual number you can work it out to be 547.32

Using one of the rules of logs - x^(logx(y))= y

You can rewrite the log as - e^(log e^(2x+2)) = e^7 -- to isolate whats in the bracket (2x+2)

You'll now be left with - 2x+2=e^7

Solving for X 2x=e^7-2

x=(e^7-2)/2

Because e is an actual number you can work it out to be 547.32

Guest Sep 27, 2013

#2**0 **

I'm sorry, I don't get what Scott is saying although it would be easier to follow if this site allowed maths type.

Any way the question can only be answered if the original log is base e. So I will assume that.

log e^x = xloge = x*1 = x

So log (base e) of e^(2x+2) = 2x+2

The problem becomes 2x+2 = 7 The answer is x = 2.5

Any way the question can only be answered if the original log is base e. So I will assume that.

log e^x = xloge = x*1 = x

So log (base e) of e^(2x+2) = 2x+2

The problem becomes 2x+2 = 7 The answer is x = 2.5

Melody Sep 28, 2013