SOLUTION: I'm finding the inverse of f -> f^1 In the process given, it says: x = log(y+2)-6 x+6 = log(y+2) //Isolate the logarithm 10^x+6= y+2 //Change to an exponential for

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: I'm finding the inverse of f -> f^1 In the process given, it says: x = log(y+2)-6 x+6 = log(y+2) //Isolate the logarithm 10^x+6= y+2 //Change to an exponential for      Log On


   

Question 623775: I'm finding the inverse of f -> f^1
In the process given, it says:
x = log(y+2)-6
x+6 = log(y+2) //Isolate the logarithm
10^x+6= y+2 //Change to an exponential form
Solve for Y

Where I'm stuck in the actual work is the 'change to exponential form' part. Can someone breakdown the process by which the example arrives at 10^x+6 from the x+6 = log(y+2) ?
Thank you.
Answer by math-vortex(648) About Me  (Show Source):
You can put this solution on YOUR website!
Hi, there--

The Problem: Solve for y.
x = log(y+2)-6 

Add 6 to both sides in order to isolate the logarithmic expression
x+6 = log(y+2)

When I get ready to convert a logarithmic equation to an exponential equation, I always say,
LOGARITHMIC: "Log to the BASE ____  of a NUMBER equals an EXPONENT."
EXPONENTIAL: "A BASE ____ raised to an EXPONENT equals a NUMBER."

In your logarithmic equation, you have "Log to the base 10 of (y+2) equals x+6." Whenever no 
base is shown (the little subscript right after the word "log), the base is understood to be 10.
BASE = 10
NUMBER = y+2
EXPONENT = x+6

Translating to exponential form, we have base 10 raised to "x+6" equals "y+2".
10^(x+6) = y+2 

Subtract 2 from both sides to solve for y.
y=10%5E%28x%2B6%29%2B2

I hope this helps! Please email me if you have questions about any part of this. I'll be happy to help 
you sort it out.

Ms.Figgy
math.in.the.vortex@gmail.com