Posted by **kelly** on Sunday, June 12, 2011 at 10:13pm.

solve each equation algebraically and check it by substituing into the orignall equation.

50e^0.035x=200

3LN(x-3)+4=5

Method of your choice by solving.

logx^2=6

Logx^4=2

2x-2^-x/2=4

e^x+e^-x/2=4

500/1+25e^.3x=200

- pre calculous -
**Reiny**, Sunday, June 12, 2011 at 11:35pm
50e^(0.035x)=200

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

by definition:

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

e^x+e^-x/2=4

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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