Posted by kelly on Sunday, June 12, 2011 at 10:13pm.
solve each equation algebraically and check it by substituing into the orignall equation.
Method of your choice by solving.
pre calculous - Reiny, Sunday, June 12, 2011 at 11:35pm
e^(.035x) = 4
take ln of both sides
ln(e^(.035x)) = ln4
.035x lne = ln4, but lne = 1
x = ln4 / .035 = appr. 39.6
log x^2 = 6
x^2 = 10^6
x = 10^3 = 1000
do logx^4 = 2 the same way
2x-2^-x/2=4 : I have a feeling that is not what you meant, without brackets I cannot tell what the equation is. I think the first term is probably 2^x
Again, I think you meant:
e^x+e^(-x/2) = 4
let e^(-x/2) = y , where y is a positive real number
then e^(x/2) = 1/y
and e^x = 1/y^2
so 1/y2 + y = 4
1 + y^3 = 4y^2
y^3 - 4y^2 + 1 = 0
I used a program to find
y = .54 or y = 3.94 appr. , there is also a negative root which would not be allowed
so e^(-x/2) = .54 or ..... use the other root
-x/2 = ln.54
x = 1.2324 or ......
last question, way too ambiguous without brackets.
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