anonymous
  • anonymous
solve by creating same bases: 1/9 = 27 ^ x-1
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Mertsj
  • Mertsj
\[\frac{1}{9}=3^{-2}\]
Mertsj
  • Mertsj
\[27^{x-1}=(3^{3})^{x-1}\]
Mertsj
  • Mertsj
\[(3^{3})^{x-1}=3^{3x-3}\]

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Mertsj
  • Mertsj
So:\[3^{-2}=3^{3x-3}\]
Mertsj
  • Mertsj
And -2=3x-3 1=3x 1/3=x
anonymous
  • anonymous
Can you please help me with one more??? PLEASE!! begging you.
Mertsj
  • Mertsj
Post it.
anonymous
  • anonymous
properties of logarithms: 3^x=5^x+2
anonymous
  • anonymous
3^x=5^(x+2)
Mertsj
  • Mertsj
\[x \log_{10}3=(x+2)\log_{10}5 \]
anonymous
  • anonymous
How did you get that?
Mertsj
  • Mertsj
Find the log of 3 and the log of 5 using your calculator and solve the equation.
Mertsj
  • Mertsj
\[a ^{n}=n \log_{10}a \]
anonymous
  • anonymous
how did you know what the variables were?
Mertsj
  • Mertsj
That is the property of law of logs. Perhaps your book uses different variables. What the variables are doesn't matter. Did you book say: \[m ^{n}=n \log_{10}m \]
anonymous
  • anonymous
oh eyes!
anonymous
  • anonymous
yes
Mertsj
  • Mertsj
So use that property on 3^x
Mertsj
  • Mertsj
Would you not get xlog3?
anonymous
  • anonymous
I think my answer is wrong :/
Mertsj
  • Mertsj
What did you get?
anonymous
  • anonymous
4.52
Mertsj
  • Mertsj
Hang on.
Mertsj
  • Mertsj
xlog3=.477x and (x+2)log 5 = .699x+1.40
Mertsj
  • Mertsj
do you agree with that/
anonymous
  • anonymous
Yes!
Mertsj
  • Mertsj
Subtract .699x from both sides.
Mertsj
  • Mertsj
-.222x=1.40
anonymous
  • anonymous
Oh so the final answer is 1.40?
Mertsj
  • Mertsj
Now divide both sides by -.222
anonymous
  • anonymous
Oh so -6.31?
Mertsj
  • Mertsj
yes.
anonymous
  • anonymous
THANK YOU SO MUCH! seriously, you are my life-saver. Now I have one more, I don't want to make you have to show me but could I just ask you a question about it?
Mertsj
  • Mertsj
Yes
anonymous
  • anonymous
http://gyazo.com/8020e5d281ab10e0d5802e2690ad9794
anonymous
  • anonymous
I don't get if there's acertain equation I am supposed to apply because it looks so confusing.
Mertsj
  • Mertsj
This involves two laws of logs. On the left side you have the difference of two logs which is equal to the log of the quotient.
Mertsj
  • Mertsj
\[\log_{3} \frac{x+1}{x}\]
Mertsj
  • Mertsj
The right side is the property we used in the last problem.
Mertsj
  • Mertsj
\[2\log_{3}2=\log_{3}2^{2} \]
Mertsj
  • Mertsj
So now, if the logs are equal, and the bases are the same, the arguments must be equal
Mertsj
  • Mertsj
\[\frac{x+1}{x}=2^{2}\]
Mertsj
  • Mertsj
solve that equation.
Mertsj
  • Mertsj
Cross multiply
Mertsj
  • Mertsj
4x=x+1
Mertsj
  • Mertsj
3x=1 x=1/3
anonymous
  • anonymous
thank you so so so much!

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