Question

If log2(x2 – 4x + 7) = 2, find the value of x​

Answer 1

${\underline{\sf{Given}}}$

$\sf\:\log_{2}(x{}^{2}-4x+7)=2$

${\underline{\sf{To\:Find}}}$

The value of x

${\underline{\sf{Solution}}}$

$\sf\:\log_{2}(x{}^{2}-4x+7)=2$

We know that if $\sf\:\log_{b}(a) = x$ ,in exponent form $\implies \: b {}^{x} = a$

Thus,$\sf\:\log_{x{}^{2}-4x+7}=2$

$\implies\:\sf\:x{}^{2}-4x+7=4$

$\implies\:\sf\:x{}^{2}-4x+3=0$

$\implies\:\sf\:x{}^{2}-3x-x+3=0$

$\implies\:\sf\:x(x-3)-1+(x-3)=0$

$\implies\:\sf\:(x-3)(x-1)=0$

$\implies\:\sf\:x=3\:or\:x=1$

Therefore, the value of x = 3 or 1

$\rule{200}2$

More About Logarithm :

The Logarithm function is defined as

$\sf\:f(x) =\log_{b}(x)$

where b > 0 and b ≠ 1 and also x >0, reads as log base b of x.