Question

Solve for a:
2^a/ln2-2a=1/ln2

Answer 1

\[\frac{2^a}{\ln2}-2a=\frac1{\ln2}\]

Answer 2

Combine the first two terms first

Answer 3

How?

Answer 4

Multiply the second term by \(\frac{\ln2}{\ln2}\)

Answer 5

so (2^a-lna(a))/ln2

Answer 6

sorry- ln2 instead of lna

Answer 7

Yes

Answer 8

Then combine the first two terms

Answer 9

Actually you can multiply the whole thing by \(\ln2\)

Answer 10

So multiply the whole thing by ln2

Answer 11

so \[2^a-\ln2(a)=1\]

Answer 12

Yep

Answer 13

make that \[2^a-2\ln2(a)=1\]

Answer 14

i forgot the 2..

Answer 15

oh I didn't notice too sorry :P

Answer 16

now what?

Answer 17

Would you please make the a in front of the ln 2

Answer 18

sure: \[2^a-2a(\ln2)=1\]

Answer 19

Well sorry I need to think for a while :P

Answer 20

that's what happened to me.. I'm really clueless when it comes to this stuff.. transcendentals are not my thing.

Answer 21

It's not really solvable, double-check your question perhaps?

Answer 22

no, it's the right equation... can you perhaps complete the square with it?

Answer 23

http://www.wolframalpha.com/input/?i=2%5Ea%2Fln2-2a%3D1%2Fln2

Answer 24

One solution is a=0

Answer 25

The another is a = (-2 W_(-1)(-1/(2 sqrt(e)))-1)/(2 log(2))

Answer 26

I've never met the function W

Answer 27

Calculus fail.. basically just plug one side into the calculator and the other side and see where they intersect?

Answer 28

Please triple-check your question for any missing parentheses

Answer 29

do you want the original?

Answer 30

Yes

Answer 31

Find a value a such that the average value of the function f(x)=\[2^x-1\] on the interval [0,a] is equal to one.

Answer 32

That means \(\Large\displaystyle\int^a_02^x-1dx=1\)?

Answer 33

but the integral must be divided by a as that is the rule for averaage value- area divided by length of interval gives average value

Answer 34

oh sorry

Answer 35

That means \(\displaystyle\Large\frac1a\int^a_0(2^x-1)dx=1\)?

Answer 36

right

Answer 37

and the anti-derivative should be \[(2^x)/\ln2-x\]

Answer 38

2^0 is not 0 !!!

Answer 39

I know, it's 1

Answer 40

therefore, you get -(1/ln2)

Answer 41

Where did the 1 go?

Answer 42

not sure..good catch

Answer 43

wait, it's still there..

Answer 44

2^a/ln2-2a=1/ln2+1

Answer 45

not +1 because x=0

Answer 46

The average is 1

Answer 47

By the fundamental theorem of calculus, that x should change to 0, because 0 is the lower limit of integration.

Answer 48

(2^a/ln2-2a)-(1/ln2-0) = 1

Answer 49

yes, and back to where we started... unfortunately...

Answer 50

Therefore 2^a/ln2-2a=1/ln2+1

Answer 51

remember that we have to divide the entire LH side by a... and that should be just a, not 2a, by the FTC

Answer 52

Actually I'm sick so my brain is not functioning properly

Answer 53

So 2^a/ln2-a=1/ln2+1

Answer 54

change the 1 on the RH side to a and you got it...

Answer 55

Wait where did the 1/a go

Answer 56

http://www.wolframalpha.com/input/?i=solve+1%2Fa+int_0%5Ea+%282%5Ex-1%29+dx+%3D+1

Answer 57

there never was one.. the integral needed to be divided by a, and then you multiply both side by a to get it out of the denominator, changing the RH side to 1(a), or a

Answer 58

Oh

Answer 59

So your original question was correct sorry

Answer 60

yeah.. I tihnk I found a way to do it on the calculator though.... I believe you must somehow use quadratics..