SOLUTION: I am trying to review for a final and would like to get help with A &B:
a. y= (10^x-10^-x)/ (2)
b. y= (e^x +e^-x) / (2)
Algebra.Com
Question 117795This question is from textbook Fund of Alg and Trig
: I am trying to review for a final and would like to get help with A &B:
a. y= (10^x-10^-x)/ (2)
b. y= (e^x +e^-x) / (2)
This question is from textbook Fund of Alg and Trig
Found 2 solutions by stanbon, checkley71:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
a. y= (10^x-10^-x)/ (2)
Multiply both sides by 2 to get:
2y = (10^x -10^-x)
Multiply both sides by 10^x to get:
10^x(2y) = 10^2x-1
Rearrange:
[10^x]^2 - [10^x](2y) + 1 = 0
Use the Quadratic Formula:
10^x = [2y +- sqrt((2y)^2-4*1*1)]/2
Simplify:
10^x = [2y +- sqrt(4y^2-4)]/2
10^x = y +- sqrt(y^2-1)
------------
Solve for "x" : x = log[y +- sqrt(y^2-1)]
Solve for "y" : y = 10^x + sqrt(y^2-1)
------------
---------------
b. y= (e^x +e^-x) / (2)
2y = (e^x + 1/e^x)
Multiply thru by e^x to get:
(2y)e^x = e^2x + 1
Rearrange:
e^2x -(2y)^x + 1 = 0
Use Quadratic Formula:
e^x = [2y +- sqrt((2y)^2-4*1*!)]/2
e^x = [2y +- sqrt(4y^2-4)]/2
e^x = [y +- sqrt(y^2-1)]
Solve for x: x = ln[y +- sqr(y^2-1)]
Solve for y: y = e^x + sqrt(y^2-1)
=======================
Cheers,
Stan H.
Answer by checkley71(8403) (Show Source): You can put this solution on YOUR website!
I AM NOT FAMILIAR WITH THE TECHNIQUE YOU MENTIONED. SORRY.
------------------------------------------------------------
A) Y=(10^X-10^-X)/2
Y=(10^0)/2
Y=1/2 ANSWER.
B) Y=(e^X+e^-X)/2
Y=(e^0)/2
Y=1/2 ANSWER.
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