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

 Dec 12, 2019
 #1
avatar+21725 
0

As someone said before, by inspection, x = 81

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

 Dec 12, 2019
 #2
avatar+21725 
0

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

 

log10 x / log10 3     log10 x / log10 9  + log10 x / log10 81 = 7

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

3.66783 log x = 7

log x = 1.9084

x = 81

ElectricPavlov  Dec 12, 2019
 #3
avatar+29200 
+5

Could also do it like this:

 

Let a=log3x.     b=log9x.      c=log81x

 

Then x=3a    x=9b=32b   x=81c=34c

  

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

 Dec 12, 2019

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