SOLUTION: I have 2 questions: 1. {{{y=5^(x)}}} {{{y=5^(2x) -1}}} 2. {{{y=2e^(2x)}}} {{{y=e^(x) -1}}}

Algebra ->  Systems-of-equations -> SOLUTION: I have 2 questions: 1. {{{y=5^(x)}}} {{{y=5^(2x) -1}}} 2. {{{y=2e^(2x)}}} {{{y=e^(x) -1}}}      Log On


   

Question 852351: I have 2 questions:
1. y=5%5E%28x%29
y=5%5E%282x%29+-1
2. y=2e%5E%282x%29
y=e%5E%28x%29+-1
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
The first one seems potentially quadratic in form, when adjusted right.
5%5Ex=5%5E%282x%29-1
5%5E%282x%29-5%5Ex-1=0
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Let u=5%5Ex. This makes 5%5E%28x%29%2A5%5Ex=5%5E%282x%29=u%5E2.
Substituting you make
u%5E2-u-1=0
u=%281%2B-+sqrt%281-4%28-1%29%29%29%2F2
u=%281-sqrt%285%29%29%2F2 or u=%281%2Bsqrt%285%29%29%2F2
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Back-substituting from those two solutions for u,
NEG SQRT VERSION
5%5Ex=%281-sqrt%285%29%29%2F2-----the right member would become negative, and you cannot raise 5 to any power to get a negative number. Reject This Equation.
POS SQRT VERSION
5%5Ex=%281%2Bsqrt%285%29%29%2F2-----THIS we can use. Irrational.
5%5Ex=1.61803....
Choose your logarithm base. Here, using base ten:
log%2810%2C5%5Ex%29=log%2810%2C1.61803%29
x%2Alog%2810%2C5%29=0.20899
x=%280.20899%29%2Flog%2810%2C5%29
highlight%28x=0.2990%29