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

Question

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

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Answer 1 · 2017-04-13T11:56:45

#x=1#

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 · 2017-04-13T12:20:35

#x=1#

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#