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Have to solve this algebraic. I end up with the wrong result. I think I'm messing up with the log rules or something. Help please

$ 2 \cdot 1,3^x = 12 \cdot 0,9^x $

$ 1,3^x = 6 \cdot 0,9^x $

$ x \cdot \log(1,3) = x \cdot \log (6 \cdot 0,9 )$

Raffaele
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Amy A
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  • That is not a correct application of the properties of the logarithm on the RHS. On that side, the exponent $x$ applies only to $0.9$. – Fimpellizzeri Mar 05 '21 at 19:13
  • So third line is: x * log(1.3) = x * log(0.9) * 6 ? – Amy A Mar 05 '21 at 19:39
  • No. The third line should be $\log\left(1.3^x\right) = \log\left(6\cdot0.9^x\right)$. Then proceed by applying properties of the logarithm individually to each side of the equality. – Fimpellizzeri Mar 05 '21 at 19:42
  • But shouldn't I move the x outside the parenthesis and multiply with the log thing? Is it: x * log(1.3) = x * log(6) * log (0.9) ? – Amy A Mar 05 '21 at 19:49
  • There is a property of the logarithm which is $\log(a^x) = x\log(a)$. However, on the RHS, your expression has the form $\log(b\cdot a^x)$, so you need to do something else first. – Fimpellizzeri Mar 05 '21 at 19:53
  • But isn't that what I've done? RHS: log(ba^x) = x log (a * b) = x * (6 * 0.9) – Amy A Mar 05 '21 at 19:59
  • The property does not apply when there is something multiplying $a^x$ inside the logarithm. – Fimpellizzeri Mar 05 '21 at 20:05
  • Damn, didn't know that. So just RHS: log(6*0.9^x)? – Amy A Mar 05 '21 at 20:10

1 Answers1

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$ 2 \cdot 1,3^x = 12 \cdot 0,9^x $

$ 1,3^x = 6 \cdot 0,9^x $

$ x \log 1,3 = \log 6+ x\log 0,9$

$x(\log 1,3-\log 0,9) =\log 6$

$$x=\frac{\log 6}{\log 1,3-\log 0,9}$$

Raffaele
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