Question 1210406
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8^x = 16^x + 4
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<pre>
I read this equation exactly and literally as it is written

    {{{8^x}}} = {{{16^x}}} + {{{4}}}.     (1)



My statement is that this equation has no solutions in real numbers.



Indeed, if  x >= 0,  then, obviously,  {{{8^x}}}  is less than  {{{16^x}}};

Hence, a fortiori,  {{{8^x}}} < {{{16^x}}} + {{{4}}},  so equation (1) has no solutions in the domain  x >= 0.



Next, if  x < 0,  then left side of equation (1) is less than 1,
while right side of this equation is greater than 4, so equation (1) has no solutions in the domain   x < 0.



Thus the statement is proved and the equation has no solutions in real domain.
</pre>

Solved.



<pre>
If, in opposite, your original equation is

    {{{8^x}}} = {{{16^(x+4)}}},    (2)


then the reasoning is different and the answer is different, too.


Then this equation (2) can be rewritten equivalently in base '2'

    {{{2^(3x)}}} = {{{2^(4*(x+4))}}},


which implies an equation for indexes

    3x = 4*(x+4),

    3x = 4x + 16,

    3x - 4x = 16,

       -x   = 16,

        x   = -16,


so the solution for equation (2) is x = -16.
</pre>

Solved two times for two different interpretations.


By the way, both interpretations lead to meaningful and instructive solutions.