Question

please answer me this question​

Answer 1

Solution :

$\tt log_x \Bigg[log_4 \bigg[log_x \bigg(5 {x}^{2} + 4 {x}^{3} \bigg) \bigg] \Bigg] = 0$

$\implies \tt {x}^{0 } = log_4 \bigg[log_x \bigg(5 {x}^{2} + 4 {x}^{3} \bigg) \bigg]$

[ Because logₐN = x ⇔ a^x = N ]

$\implies \tt 1 = log_4 \bigg[log_x \bigg(5 {x}^{2} + 4 {x}^{3} \bigg) \bigg]$

[ Because a^0 = 1 ]

$\implies \tt log_44 = log_4 \bigg[log_x \bigg(5 {x}^{2} + 4 {x}^{3} \bigg) \bigg]$

[ Because logₐa = 1 ]

Comparing on both sides

$\implies \tt 4 = log_x \bigg(5 {x}^{2} + 4 {x}^{3} \bigg)$

$\implies \tt log_x \bigg(5 {x}^{2} + 4 {x}^{3} \bigg) = 4$

$\implies \tt {x}^{4} = 5 {x}^{2} + 4 {x}^{3}$

[ Because logₐN = x ⇔ a^x = N ]

$\implies \tt {x}^{4} - 4 {x}^{3} - 5 {x}^{2} = 0$

$\implies \tt {x}^{2} ( {x}^{2} - 4x - 5) = 0$

$\implies \tt {x}^{2} - 4x - 5= 0$

Splitting the middle term

$\implies \tt {x}^{2} - 5x + x - 5= 0$

$\implies \tt x(x - 5) + 1(x - 5) = 0$

$\implies \tt (x + 1)(x - 5) = 0$

$\implies \tt x + 1 = 0 \quad or \quad x - 5 = 0$

$\implies \tt x = - 1 \quad or \quad x = 5$

[ Negleting x = - 1 because bases of logarithms cannot be negative and 0 ]

$\implies \tt \boxed{x = 5}$

Hence, the value of x is 5.

Answer 2

Question :-

$\sf{ log_{x} \bigg( log_{4} \big( log_{x}(5 {x}^{2} + 4 {x}^{3} ) \big) \bigg) = 0}$

Answer :-

→ x = 5

Solution :-

According to the question,

$\sf{ log_{x} \bigg( log_{4} \big( log_{x}(5 {x}^{2} + 4 {x}^{3} ) \big) \bigg) = 0} \\ \: \\ \because \sf{ log_{n}(s) = p \: \rightarrow \: s = {n}^{p} } \\ \therefore \\ \rightarrow \sf{ log_{4} \bigg( log_{x}(5 {x}^{2} + 4 {x}^{3} ) \bigg) \: = {x}^{0} } \\ \\ \rightarrow \sf{ log_{4} \bigg( log_{x}(5 {x}^{2} + 4 {x}^{3} ) \bigg) \: = 1} \\ \\ \rightarrow \sf{ log_{x}(5 {x}^{2} + 4 {x}^{3} ) = {4}^{1} } \\ \\ \rightarrow \sf{5 {x}^{2} + 4 {x}^{3} = {x}^{4} } \\ \\ \rightarrow \sf{ {x}^{4} - 4 {x}^{3} - 5 {x}^{2} = 0} \\ \\ \rightarrow \sf{ {x}^{2} ( {x}^{2} - 4x - 5) = 0} \\ \\ \star \: \text{case \: (1)} \\ \implies \boxed{ \sf{x} = 0 }\\ \: \\ \star \: \text{case \: (2)} \\ \\ \implies \sf{ {x}^{2} - 4x - 5 = 0} \\ \\ \implies \sf{ {x}^{2} - 5x + x - 5 = 0} \\ \\ \implies \sf{x(x - 5) + 1(x - 5) = 0} \\ \\ \implies \sf{ (x - 5)(x + 1) = 0} \\ \\ \rightarrow \boxed{ \sf{x \: = 5 \: or \: - 1}}$

But here negative base of log is not defined therefore x = 5

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