Can you help me find x? (Exponential Function).

2^x -10 (2^-x)+ 3 = 0

2 Answers
Shwetank Mauria
Apr 13, 2017

x=1

Explanation:

Multiplying each term of 2^x-10(2^-x)+3=0 by 2^x, we gwt

2^(2x)-10+3xx2^x=0.

Now let 2^x=u, the above becomes

u^2+3u-10=0

or u^2+5u-2u-10=0

or u(u+5)-2(u+5)=0

or (u-2)(u+5)=0

but as u=2^x, u+5!=0 and dividing both sides by it, we get

u=2 or 2^x=2 or x=1

Tony B
Apr 13, 2017

x=1

Explanation:

Write as:" "2^x-10/2^x+3" "=" "0

" "color(green)([2^xcolor(red)(xx1)] -10/2^x+[3color(red)(xx1)]" "=" "0

" "color(green)([2^xcolor(red)(xx2^x/2^x)] -10/2^x+[3color(red)(xx2^x/2^x)]=0

" "(2^(2x))/2^x" " -10/2^x+" "(3xx2^x)/2^x" "=" "0

" "2^(2x)+(3xx2^x)-10=0

Changed my mind about the format. Write 2^(2x) " as "(2^x)^2

Set a=2^x giving

a^2+3a-10=0

(a-2)(a+5)=0

a=-2 and a=+5

=>2^x=+2 and 2^x=-5

Using logs

Consider 2^x=2

xln(2)=ln(2) => x=ln(2)/ln(2)=1

Consider 2^x=-5

xln(2)=ln(-5)

but ln(-5) is undefined so not a solution.

Thus x=1

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