SOLUTION: Use the one to one property to solve the equations for x A) log 11=log (x^2+7) B)e ^x^2-3=e^2x

Algebra ->  Logarithm Solvers, Trainers and Word Problems -> SOLUTION: Use the one to one property to solve the equations for x A) log 11=log (x^2+7) B)e ^x^2-3=e^2x      Log On


   

Question 1136697: Use the one to one property to solve the equations for x
A) log 11=log (x^2+7)
B)e ^x^2-3=e^2x
Found 2 solutions by MathLover1, Theo:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

A)
log+%2811%29=log+%28x%5E2%2B7%29...if log same then
11=x%5E2%2B7
11-7=x%5E2
x%5E2=4
x=sqrt%284%29
x± 2


B)
e+%5E%28x%5E2-3%29=e%5E%282x%29...if base same then exponents are same too
+x%5E2-3=2x
+x%5E2-2x-3=0
+x%5E2%2Bx-3x-3=0
+%28x%5E2%2Bx%29-%283x%2B3%29=0
+x%28x%2B1%29-3%28x%2B1%29=0
+%28x-3%29%28x%2B1%29=0
solutions: x=3 or x=-1






Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
a.)

log(11) = log(x^2 + 7)

this is true if and only if 11 = x^2 + 7

subtract 7 from both sides to get 4 = x^2.

solve for x to get x = plus or minus 2.

when x = plus or minus 2, x^2 is 4.

log(11) = log(x^2 + 7) becomes log(11) = log(4 + 7) which becomes log(11) = log(11) which confirms the solution is good.

b.)

e^(x^2-3) = e^(2x)

this is true if and only if x^2 - 3 = 2x

subtract 2x from both sides to get x^2 - 2x - 3 = 0

factor to get (x-3) * (x+1) = 0

solve for x to get x = -1 or x = 3.

when x = -1, e^(x^2-3) = e^(2x) becomes e^(1-3) = e^(2*-1) which becomes e^(-2) = e^(-2) which confirms the solution is good.

when x = 3, e^(x^2-3) = e^(2x) becomes e^(9-3) = e^(2*3) which becomes e^(6) = e^(6) which confirms the solution is good.