How to solve these question without using a calculator?

2012-01-02 5:34 pm
e^2x-2e^x-3=0
In(2x+1)=10

回答 (4)

2017-01-09 7:07 pm
are you able to apply a working laptop or computing gadget? often times you may guess the respond, if this is small and not having decimal places. alongside with your equation, i do no longer think of this is available (it has loads of decimal places). whether this is: x^4 + 5x^3 + 5x^2 - 5x - 6 = 0 this is available, because of the fact the coefficients upload as much as 0, so one answer may be x=a million eqn=x^4 - x^3 + 6x^3 - 6x^2 + 11x^2 - 11x + 6x - 6 = 0 (x-a million)(x^3 + 6x^2 + 11x + 6) = 0 the coefficients of even powers = coefficients of wierd powers, so one answer may be x=-a million (x-a million)(x^3 + x^2 + 5x^2 + 5x + 6x + 6) = 0 (x-a million)(x+a million)(x^2 + 5x + 6) = 0 relatively see the final solutions are x = -2 and x = -3 (x-a million)(x+a million)(x+2)(x+3)=0 you additionally can try with x^4 + 5x^3 - 5x^2 + 5x - 6 = 0, yet YOUR eqn, no.
2012-01-02 5:50 pm
for the first one

e^2x-2e^x-3=0
you effectively have a quadratic equation of e^x

so you can write down

(e^x)^2 -2(e^x) -3 =0

it can then be factorised into

((e^x)+1)((e^x)-3) = 0

therfore either

e^x=-1 you can never get a negative number from an exponential so this has no solutions but

e^x = 3

therefore x=ln3 I think you should be able to leave it in this form if you don't have a calculator

for your second question

ln(2x+1)=10

to get rid of the ln you take the exponential of both sides and you get

2x+1 = e^10

then you just rearrange the equation to get x= ((e^10) -1)/2

I hope this helps
2012-01-02 5:48 pm
e^(2x) - 2e^x - 3 = 0

let e^x = a, ==> e^(2x) = a^2

=> a^2 - 2a - 3 = 0

=> (a - 3)(a + 1) = 0

a = -1 and 3

=> e^x = 3 (ignoring negative value since e^x can't be negative )

x = ln(3)

2)

ln(2x + 1) = 10

2x + 1 = e^10

2x = e^(10) - 1

x = 1/2 [e^10 - 1 ]
2012-01-02 5:43 pm
e^2x-2e^x-3 = (e^x)^2 -2e^x - 3

now we can factor it:

(e^x)^2 -2e^x - 3 = (e^x + 1)(e^x - 3) = 0

This happens when

e^x = -1 or
e^x = 3

e^x never equals -1 because only positive reals are in its range. So it's only true when

e^x = 3
log both sides.

x = ln3

This is as far as you can get without a calculator.

Now this one

In(2x+1)=10

raise e to the power of both sides

2x + 1 = e^10

x = (e^10 - 1)/2

Once again, this is as far as you can get without a calculator.


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