<s>[INST] How would i go about solving this exponential equation? $$2^{ x-2 }+\frac { 1 }{ \sqrt { { 4 }^{ x+2 } }  } =\frac { 17 }{ 16 } $$
I can't seem to find any direction to go that will lead me to a solution. I decided to try to take the natural log all terms in the equation to see where that would lead me. Please do not give me the solution. I just want to be put in the right direction so that I could get the solution on my own. Thanks.
This is what I have done so far:
$$ln(2^{ x-2 })+ln(\frac { 1 }{ \sqrt { { 4 }^{ x+2 } }  } )=ln(\frac { 17 }{ 16 } )$$
$$(x-2)ln(2)+[ln(1)-ln(\sqrt { { 4 }^{ x+2 } } )]=ln(17)-ln(16)$$
$$xln(2)-2ln(2)+[ln(1)-ln(4\sqrt { { 4 }^{ x } } )]=ln(17)-ln(16)$$
 [/INST] There's no need to use logarithms at all here-just indices will do.
Note that: $$\frac{1}{\sqrt{4^{x+2}}}= \frac{1}{2^{x+2}} = \frac{1}{4(2^x)}$$
$$ 2^{x-2} + \frac{1}{2^{x+2}} = \frac{17}{16} $$
$$ \frac{2^x}{4}+ \frac{1}{2^x(4)}= \frac{17}{16}$$
Let $2^x = a$
$$\frac{a}{4}+ \frac{1}{4a}= \frac{17}{16}$$
$$ 4a^2 -17a +4 = 0$$
$$a = 2^{-2}\  \text{or} \ 4 $$
Now substitute the values of $a$ into $2^x = a$ to get the values of $x$.
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