Question 1103126
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<pre>
Your equation is

{{{2^x - 10*2^(-x) + 3}}} = 0.


It is a quadratic equation (or EQUIVALENT to a quadratic equation) relative to the variable {{{2^x}}}.


To see it CLEARLY, introduce new variable u = {{{2^x}}}.


Then your equation takes the form

{{{u - 10*(1/u) + 3}}} = 0.


Multiply by u to rid off the denominator. You will get

{{{u^2 - 10 + 3u}}} = 0,

{{{u^2 + 3u - 10}}} = 0.


Factor left side

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


Only postive root works and make sense: u = 2.


It means {{{2^x}}} = u = 2,  which implies  x = 1.
</pre>

Solved.


That's all.


Introducing new variable is the STANDARD method of solutions to problems like this.


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See the lesson

&nbsp;&nbsp;&nbsp;&nbsp;- <A HREF=https://www.algebra.com/algebra/homework/logarithm/How-to-solve-exponential-equations.lesson>Solving exponential equations</A>

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