Question

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

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

Answer 1

`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`

Answer 2

`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`