Question

log x base 2 + log (x+1) base 2=1
do this please​

Answer 1

Question :

Solve for x :

$\tt {log}_{2}{x} + {log}_{2}{(x+1)}$ = 1

Solution :

$\tt {log}_{2}{x} + {log}_{2}{(x+1)}$ = 1

$\tt \implies {log}_{2}{x(x+1)} = {log}_{2}{2}$

$\tt \implies$ x(x+1) = 2

$\tt \implies {x}^{2}$ + x = 2

$\tt \implies {x}^{2}$ +x - 2 = 0

$\tt \implies {x}^{2}$ + 2x - x - 2 = 0

$\tt \implies$ x(x+2) - 1(x+2) = 0

$\tt \implies$ (x+2)(x-1) = 0

$\tt \implies$ x = - 2 , 1

x = - 2 , 1

Properties of logarithm :

● $\tt {log}_{c}{a} + {log}_{c}{b} = {log}_{c}{ab}$

●$\tt {log}_{c}{a} - {log}_{c}{b} = {log}_{c}{\frac{a}{b}}$

● $\tt {log}_{c}{{a}^{b}} = b{log}_{c}{a}$

● $\tt {log}_{a}{a} = 1$

●$\tt {log}_{a}{1} = 0$