SOLUTION: Hello! I need help on this one math problem. It seems complicated but I know that it is easier than it looks. Please, if you can , help me!! Problem: Solve for x Verify your

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Hello! I need help on this one math problem. It seems complicated but I know that it is easier than it looks. Please, if you can , help me!! Problem: Solve for x Verify your       Log On


   

Question 39504: Hello! I need help on this one math problem. It seems complicated but I know that it is easier than it looks. Please, if you can , help me!!
Problem: Solve for x
Verify your result by checking that it approximately satisfies the original equation.
2^4x+5=17
log(2^4x+5)=log17
4x+5log2=log17
And that's where I stopped because I have NO IDEA what to do next!!
HELP MEEEEEEEEEEEE!!
Answer by AnlytcPhil(1807) About Me  (Show Source):
You can put this solution on YOUR website!
Hello! I need help on this one math problem.
It seems complicated but I know that it is 
easier than it looks. Please, if you can ,
help me!! 

Problem: Solve for x 

Verify your result by checking that it 
approximately satisfies the original equation. 

2^(4x+5) = 17

log[2^(4x+5)] = log17

(4x+5)log2 = log17 


And that's where I stopped because I have NO IDEA
what to do next!!  HELP MEEEEEEEEEEEE!!

-------

You let those logs trick you!  
They're just plain old numbers!!!:

Get your calculator and find

log2 = .301 and log17 = 1.23 approximately

Now 

     (4x + 5)log2 = log17 

becomes

   (4x + 5)(.301) = 1.23

or

     .301(4x + 5) = 1.23

Distribute:

   1.204x + 1.505 = 1.23

Solve for x

           1.204x = 1.23 - 1.505

           1.204x = -.275

                x = -.27/1.204

                x = -.224

Check with calculator


         2^(4x+5) = 17

2^[4·(-.224) + 5] = 17

    2^[-.896 + 5] = 17

          2^4.104 = 17

             17.2 = 17

Fairly close.  If you want it to be closer go 
back and use more decimal places.

Edwin
AnlytcPhil@aol.com