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Find the real value of x that satisfies \(\log_3 x + \log_9 x + \log_{81} x = 7\)

Guest Dec 12, 2019

ElectricPavlov · Answer 1 · 2019-12-12T02:52:33

+36916

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As someone said before, by inspection, x = 81

log(81,3)+log(81,9)+log(81,81) = 7

+36916

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a Little more detail ....how to SOLVE algebraically: (using base change log rule)

log₁₀x / log10 3 log₁₀x / log₁₀ 9 + log₁₀ x / log₁₀81 = 7

2.0959 log x + 1.04795 log x + .5239 log x = 7

3.66783 log x = 7

log x = 1.9084

x = 81

Alan · Answer 2 · 2019-12-12T04:32:39

Could also do it like this:

Let a=log₃x. b=log₉x. c=log₈₁x

Then x=3^a x=9^b=3^2b x=81^c=3^4c

So a=2b and a=4c

But a+b+c=7 so a + a/2 + a/4 = 7 or 7a/4 = 7 or a = 4

Hence x = 3⁴ = 81