Question

If 2log base 16 (x²+x) - log base 4 (x +1) find x.​

Answer 1

Correct question:

If $2 log_{16}({x}^{2}+x) = log_{4}(x+1)$ , then find x.

Answer:

$\large\boxed{\sf{x=1}}$

Step-by-step explanation:

Given a logarithmic equation such that,

$2 log_{16}( {x}^{2} + x ) = log_{4}(x + 1)$

To find the value of x,

$= > 2 log_{ {4}^{2} }( {x}^{2} + x) = log_{4}(x + 1)$

But, we know that,

• $\large{ log_{ {a}^{p} }( {m}^{q} ) = \dfrac{q}{p} log_{a}(m) }$

Therefore, we will get,

$= > \dfrac{2}{2} log_{4}( {x}^{2} + x ) = log_{4}(x + 1) \\ \\ = > log_{4}( {x}^{2} + x) = log_{4}(x + 1)$

Now, on both sides, logarithmic bases are same.

Therefore, we will get,

$= > {x}^{2} + x = x + 1 \\ \\ = > {x}^{2} - 1 = 0 \\ \\ = > (x + 1)(x - 1) = 0 \\ \\ = >\bold{ x = \pm1}$

But, on putting x = -1, the function will be invalid.

This is because in $\bold{log_{a}(m)}$, m > 0.

Hence, required value of x = 1

Answer 2

$\huge \mathfrak{solution}$

Given:

$\star \bold{2 log_{16}( {x}^{2} + x) - log_{4}(x + 1) = 0}$

To find :

•X = ?

Answer :

$\implies \: 2 log_{16}( {x}^{2} + x ) - log_{4}(x + 1) = 0 \\ \\ \implies \: 2 log_{ {4}^{2} }( {x}^{2} + x) - log_{4}(x + 1) = 0 \\ \\ \implies \: \frac{2 }{2} log_{4}( {x}^{2} + {x} ) - log_{4}(x + 1) = 0 \\ \\ \implies \cancel{ \frac{2}{2} } log_{4}( {x}^{2} + x) - log_{4}(x + 1) = 0 \\ \\ \implies \: log_{4}( {x}^{2} + {x}) - log_{4}(x + 1) = 0\\ \\ \implies \: log_{4}( \frac{ {x}^{2} + x}{x + 1} ) = 0 \\ \\ \implies \: \frac{x( \cancel{x + 1)}}{ \cancel{x + 1}} = {4}^{0} \\ \\ \implies \: x = 1$

Hence, The value of X is 1

Formula :

$\star \bold{ log_{ {x}^{a} }(y) = \frac{1}{a} log_{x}(y) }$

$\star \bold{ log(x) - log(y) = log( \frac{x}{y} ) }$