16.4x+2-16.2x+1+1=0
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Poulami Dasgupta 7 years, 11 months ago
{tex}16.4\ ^ {x+2}-16.2\ ^ {x+1}+1=0{/tex}
{tex}16(4\ ^ x\times4\ ^2)-16(2\ ^ x\times2\ ^1)+1=0{/tex}
{tex}16\times16\times4\ ^x-16\times2\times2\ ^x+1=0{/tex}
{tex}256\times4\ ^ x-32\times2\ ^x+1=0{/tex}
{tex}256\times(2\times2)\ ^ x-32\times2\ ^ x+1=0{/tex}
Let,{tex}2\ ^x=p{/tex}
{tex}256\times p\times p-32\times p+1=0{/tex}
{tex}256p\ ^2-32p+1=0{/tex}
{tex}256p\ ^2-16p-16p+1=0{/tex}
{tex}16p(16p-1)-1(16p-1)=0{/tex}
{tex}(16p-1)(16p-1)=0{/tex}
Either {tex}16p-1=0{/tex},or,{tex}16p-1=0{/tex}
{tex}p={1\over16},{1\over16}{/tex}
Now, {tex}p=2\ ^ x{/tex}
{tex}{1\over 16}={ 2\ ^ x}{/tex},
{tex}({1\over2})\ ^ 4=2\ ^ x{/tex}
{tex}(2)\ ^ {-4}=2\ ^ x{/tex}
{tex}x= -4{/tex}
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