solve by creating same bases: 1/9 = 27 ^ x-1

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solve by creating same bases: 1/9 = 27 ^ x-1

Mathematics
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\[\frac{1}{9}=3^{-2}\]
\[27^{x-1}=(3^{3})^{x-1}\]
\[(3^{3})^{x-1}=3^{3x-3}\]

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So:\[3^{-2}=3^{3x-3}\]
And -2=3x-3 1=3x 1/3=x
Can you please help me with one more??? PLEASE!! begging you.
Post it.
properties of logarithms: 3^x=5^x+2
3^x=5^(x+2)
\[x \log_{10}3=(x+2)\log_{10}5 \]
How did you get that?
Find the log of 3 and the log of 5 using your calculator and solve the equation.
\[a ^{n}=n \log_{10}a \]
how did you know what the variables were?
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: \[m ^{n}=n \log_{10}m \]
oh eyes!
yes
So use that property on 3^x
Would you not get xlog3?
I think my answer is wrong :/
What did you get?
4.52
Hang on.
xlog3=.477x and (x+2)log 5 = .699x+1.40
do you agree with that/
Yes!
Subtract .699x from both sides.
-.222x=1.40
Oh so the final answer is 1.40?
Now divide both sides by -.222
Oh so -6.31?
yes.
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?
Yes
http://gyazo.com/8020e5d281ab10e0d5802e2690ad9794
I don't get if there's acertain equation I am supposed to apply because it looks so confusing.
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.
\[\log_{3} \frac{x+1}{x}\]
The right side is the property we used in the last problem.
\[2\log_{3}2=\log_{3}2^{2} \]
So now, if the logs are equal, and the bases are the same, the arguments must be equal
\[\frac{x+1}{x}=2^{2}\]
solve that equation.
Cross multiply
4x=x+1
3x=1 x=1/3
thank you so so so much!

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