Cómo resolver log 36 a la base 16 = 1 – log x a la base de 4

[matemáticas] \ log_ {16} {36} = 1 – \ log_ {4} {x} [/ matemáticas]

Recuerde que [math] \ log_ {a} {b} = \ dfrac {\ log {b}} {\ log {a}} [/ math].

[matemáticas] \ por lo tanto \ dfrac {\ log {36}} {\ log {16}} = 1 – \ dfrac {\ log {x}} {\ log {4}} [/ math]

[matemáticas] \ por lo tanto \ log {x} = \ left (1 – \ dfrac {\ log {36}} {\ log {16}} \ right) \ times \ log {4} [/ math]

Recuerde que [math] \ log {a ^ 2} = 2 \ log {a} [/ math], entonces [math] \ log {16} = 2 \ log {4} [/ math] y [math] \ log {36} = 2 \ log {6} [/ math].

[matemáticas] \ por lo tanto \ log {x} = \ left (1 – \ dfrac {2 \ log {6}} {2 \ log {4}} \ right) \ times \ log {4} = \ log {4} – \ log {6} [/ matemáticas]

Recuerde que [math] \ log {a} – \ log {b} = \ log \ left (\ frac {a} {b} \ right) [/ math].

[matemáticas] \ por lo tanto \ log {x} = \ log \ left (\ frac {2} {3} \ right) [/ math]

[matemáticas] \ por lo tanto \ boxed {x = {\ textstyle \ frac {2} {3}}} [/ math]

Log 36 / Log 16 = 1- Log x / Log 4 =>

Log 36 / (2 Log 4) = (Log 4 – Log x) / Log 4 =>

Log 36 = 2 Log 4 -2 Log x =>

2 Log x = Log 16 – Log 36 = Log 16/36 =>

Log x ^ 2 = Log 4/9 => x ^ 2 = 4/9 => x = 2/3

log 36 a la base16 = 1-log x a la base 4

(log36) / log16 = 1- (log x) / (log 4)

(2log2 + 2log3) / (4log2) + (log x) / (2log 2) = 1

(log2 + log3) / (2log2) + (logx /) (2log2) = 1

(log2 + log3 + logx) / log 4 = 1

log 6x = log4

6x = 4

x = 4/6

x = 2/3, respuesta.

(log2 + log3 + log x) / 2log2 = 1