1

Please Help me with this I think i figured out question 1... but I get no solution... please help me start number 2 or if you can show full solution that be sick thanks.

  1. $\log_{3x}(81)=2$
  2. $\log_3(\log_x(\log_4 16))=-1$

my attempt for number 1 ! :)

$$\frac{1}{81}\log_{3x}(81)=\frac{2}{81}$$ $$\log_{3x}=2/81$$ $$10^{\large{\log_{3x}}}=10(2/81)$$ $$3x=10 (2/81)$$ $$x=10 (2/81) /3 $$

luke
  • 1,307
user73122
  • 781
  • 3
  • 20
  • 39
  • number 1 I get no solution please show your solution and help me with number 2 ; – user73122 May 14 '13 at 01:25
  • 3
    Is this from an online quiz/test? Because, it surely looks like it. – Jeel Shah May 14 '13 at 01:29
  • The mistake you are making on the first one is that you cannot factor out 81. The statement is not $\log(3x)\cdot 81$, it is $\log_{3x}(81)$, the base 3x logarithm of 81 (and not 81 times the logarithm of 3x). – Ben Norris May 14 '13 at 01:52
  • I've edited your answer to try to convert to LaTeX, our math formatting software. Please make sure I'm representing your answer correctly--some things didn't seem to make sense. – apnorton May 14 '13 at 01:55

2 Answers2

1

I make multiple use of the rule: $\log_a x=b\implies a^b=x$

The first problem is already addressed in one of the other answers, but when you see $\log_{3x}(81)=2$, you should think $(3x)^2=9x^2=81\implies x=3$

As for the second problem, since $4^2=16$, we know that $\log_4 16=2$ and we can make the following simplification:

$$\log_3(\log_x(\log_4 16))=-1\\ \log_3(\log_x(2))=-1$$

Now, if we think of $\log_x(2)$ as a variable, call it $y$, we can rewrite the above equality like so: $$\log_3 y=-1\\ y=3^{-1}=\frac{1}{3}$$ Now that we have a value for $y$, lets put it back in context: $$y=\log_x(2)=\frac{1}{3}\\ x^{\frac{1}{3}}=2\\ x=8 $$

luke
  • 1,307
  • +1 for clear these thing from basic approach.I mention some of the other basic log properties in this link....http://math.stackexchange.com/questions/391655/write-the-expressoin-in-terms-of-log-x-and-log-y-log-fracx310y/391680#391680 – iostream007 May 15 '13 at 12:09
0

$$\log_{3x}(81)=2$$ $$(3x)^{\log_{3x}(81)}=(3x)^{2}$$ $$81=9x^2$$ $$9=x^2$$ $$3=x$$

russCam
  • 117