Question 1210448
.
Solve 24(2²ˣ) - 5(2ˣ⁺²) - 156 = 0, where x ∈ R
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~


<pre>
This given equation

    {{{24*2^(2x) }}} -  {{{5*2^(x+2)}}}  - 156  =  0    (1)


is the same as

    {{{24^(2x)}}}  -  {{{5*4*2^x}}}  -  156  =  0,

or

    {{{24*2^(2x)}}} - {{{ 20*2^x }}} -  156  =  0.    (2)


Introduce new variable  y = {{{2^x}}}.


Then equation (2)  takes the form

    {{{24y^2}}} - {{{20y}}} - 156 = 0,

and we are looking for positive solutions to this equation.


Apply the quadratic formula and get the solutions to equation (2)

    y = 3  and  y = -13/6.


Only positive solution y = 3 does fit.


So, for x we have

    {{{2^x}}} = 3,   hence,  x = {{{log(2,(3))}}} = 1.584962501  (approximately).


<U>ANSWER</U>.  x = {{{log(2,(3))}}} = 1.584962501  (approximately).
</pre>

Solved.


This is a standard problem to solve exponential equation by introducing new variable.