Question

solve for x.
$2 log_{10}(x) = 1 + log_{10}(x + \binom{11 \div 10}{} )$

Answer 1

Using below rules :

log 10 base 10 = 1

m*logn = log(m^n)

log a + log b = log (ab)

log(x^2) = log10 + log(x+ (11/10))

x^2 = 10(x+(11/10))

x^2 - 10x - 11 = 0

Solving for x, we get. x = 11,-1

Since log(-1) doesn't exist, x = 11