Question

Log 3 ^ x + log 9 ^ x + log 81 ^ x = 7/4

Answer 1

Answer:

$Value \: of \: x = \frac{1}{4\:log 3}$

Step-by-step explanation:

$Given \:log\:3^{x} + log\:9^{x} + log\:81^{x} =\frac{7}{4}$

$\implies log\:3^{x} + log\:(3^{2})^{x} + log\:(3^{4})^{x} =\frac{7}{4}$

$\implies log\:3^{x} + log\:3^{2x} + log\:3^{4x} =\frac{7}{4}$

$We\: know \: that ,\: \boxed { (a^{m})^{n} = a^{mn}}$

$\implies xlog\:3+2x log\:3+ 4x log\:3 =\frac{7}{4}$

$We\: know \: that ,\: \boxed { log\:n^{m} = m\:log\:n}$

$\implies (x+2x+4x)log3= \frac{7}{4}$

$\implies 7x log 3 = \frac{7}{4}$

$\implies x = \frac{7}{4}\times \frac{1}{7\:log3}$

$\implies x = \frac{1}{4\:log 3}$

Therefore.,

$Value \: of \: x = \frac{1}{4\:log 3}$

•••♪

Answer 2

Step-by-step explanation:

Step-by-step explanation:

Given \:log\:3^{x} + log\:9^{x} + log\:81^{x} =\frac{7}{4}Givenlog3x+log9x+log81x=47

\implies log\:3^{x} + log\:(3^{2})^{x} + log\:(3^{4})^{x} =\frac{7}{4}⟹log3x+log(32)x+log(34)x=47

\implies log\:3^{x} + log\:3^{2x} + log\:3^{4x} =\frac{7}{4}⟹log3x+log32x+log34x=47

We\: know \: that ,\: \boxed { (a^{m})^{n} = a^{mn}}Weknowthat,(am)n=amn

\implies xlog\:3+2x log\:3+ 4x log\:3 =\frac{7}{4}⟹xlog3+2xlog3+4xlog3=47

We\: know \: that ,\: \boxed { log\:n^{m} = m\:log\:n}Weknowthat,lognm=mlogn

\implies (x+2x+4x)log3= \frac{7}{4}⟹(x+2x+4x)log3=47

\implies 7x log 3 = \frac{7}{4}⟹7xlog3=47

\implies x = \frac{7}{4}\times \frac{1}{7\:log3}⟹x=47×7log31

\implies x = \frac{1}{4\:log 3}⟹x=4log31

Therefore.,

Value \: of \: x = \frac{1}{4\:log 3}Valueofx=4log3