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 July 26th, 2014, 09:18 PM #1 Newbie   Joined: Jul 2014 From: Seattle Posts: 16 Thanks: 0 Exponent How would I go about solving the following problem: [3 x 5^(2-x)] = 4 Thank you all!! Bella
 July 26th, 2014, 09:28 PM #2 Math Team   Joined: Dec 2013 From: Colombia Posts: 7,638 Thanks: 2623 Math Focus: Mainly analysis and algebra You will need to take logarithms to get the $x$ out of the exponent.
July 27th, 2014, 01:21 AM   #3
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Quote:
 Originally Posted by bml1105 [3 x 5^(2-x)] = 4

$\Large \color{blue}{3\cdot5^{2-x}=4 \Longleftrightarrow\ 5^{2-x} = \dfrac{4}{3}\ \Longleftrightarrow\ log_5 5^{2-x} = log_5 \dfrac{4}{3}\ \Longleftrightarrow (2-x)log_5 5 = log_5 \dfrac{4}{3} \Longleftrightarrow \\\;\\ \Longleftrightarrow\ 2-x = log_5 \dfrac{4}{3}\ \Longleftrightarrow\ 2-\log_5 \dfrac{4}{3} = x\ \Longleftrightarrow\ x = 2-\log_5 \dfrac{4}{3} .}$

Last edited by aurel5; July 27th, 2014 at 01:24 AM.

July 27th, 2014, 02:00 AM   #4
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Quote:
 Originally Posted by bml1105 [3 x 5^(2-x)] = 4
Please use * as multiplication sign:
3 * 5^(2-x) = 4

Rule: if x^p = y, then p = LOG(y) / LOG(x)
So (divide by 3):
5^(2-x) = 4/3
2 - x = LOG(4/3) / LOG(5)
Finish it...

July 27th, 2014, 07:42 AM   #5
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Quote:
 Originally Posted by Denis Please use * as multiplication sign: 3 * 5^(2-x) = 4 Rule: if x^p = y, then p = LOG(y) / LOG(x) So (divide by 3): 5^(2-x) = 4/3 2 - x = LOG(4/3) / LOG(5) Finish it...
$\displaystyle 2-x = \log_5 \frac{4}{3} \Rightarrow x = 2 - \log_5 \frac{4}{3} = \log_5 25 - \log_5 \frac{4}{3} = \log_5 \frac{25}{\frac{4}{3}}= \log_5 \frac{75}{4} = \log_5 18.75 \approx 1.82$

July 27th, 2014, 08:20 AM   #6
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Quote:
 Originally Posted by tahirimanov $\displaystyle 2-x = \log_5 \frac{4}{3} \Rightarrow x = 2 - \log_5 \frac{4}{3} = \log_5 25 - \log_5 \frac{4}{3} = \log_5 \frac{25}{\frac{4}{3}}= \log_5 \frac{75}{4} = \log_5 18.75 \approx 1.82$
You forgot the expected ending: "in other words, 5^(~1.82) = 18.75"

 July 27th, 2014, 09:11 AM #7 Newbie   Joined: Jul 2014 From: Seattle Posts: 16 Thanks: 0 Thanks everyone! I'll be sure to use a * sign next time.
July 27th, 2014, 09:29 AM   #8
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Quote:
 Originally Posted by Denis You forgot the expected ending: "in other words, 5^(~1.82) = 18.75"
I like unexpected endings....

July 27th, 2014, 10:02 AM   #9
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Quote:
 Originally Posted by aurel5 $\Large \color{blue}{3\cdot5^{2-x}=4 \Longleftrightarrow\ 5^{2-x} = \dfrac{4}{3}\ \Longleftrightarrow\ log_5 5^{2-x} = log_5 \dfrac{4}{3}\ \Longleftrightarrow (2-x)log_5 5 = log_5 \dfrac{4}{3} \Longleftrightarrow \\\;\\ \Longleftrightarrow\ 2-x = log_5 \dfrac{4}{3}\ \Longleftrightarrow\ 2-\log_5 \dfrac{4}{3} = x\ \Longleftrightarrow\ x = 2-\log_5 \dfrac{4}{3} .}$
What? No poetry?

-Dan

 July 27th, 2014, 11:17 AM #10 Newbie   Joined: Jul 2014 From: Seattle Posts: 16 Thanks: 0 Can anyone tell me how to make the subscripts for the log formulas when writing on the forum? Thanks!

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