anonymous
  • anonymous
Solve 3^(2x) = 7^(x−1).
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
hellpp
jdoe0001
  • jdoe0001
\(\bf 3^{2x} = 7^{x-1}\) right? for I don't see it happening
anonymous
  • anonymous

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jdoe0001
  • jdoe0001
hmmm
jdoe0001
  • jdoe0001
hmm hold the mayo
anonymous
  • anonymous
kk
anonymous
  • anonymous
jdoe0001
  • jdoe0001
hmmm I assume you've covered logarithms?
anonymous
  • anonymous
First step: Take the logarithm of both sides.
anonymous
  • anonymous
ao log 3^(2x)= log 7^(x-1)
anonymous
  • anonymous
hello?
anonymous
  • anonymous
OK. When you have a power, the way to take the log is as follows:\[\log a ^{b} = b \log a\]For example\[\log 5^\left( x+2 \right) = \left( x+2 \right) \log 5\] Try this with your question
anonymous
  • anonymous
3^(2x) log= 7^(x-1)= log
anonymous
  • anonymous
\[3^{2x}=7^{x-1}=7^x*7^{-1}\] \[\left( 3^2 \right)^x=\frac{ 7^x }{ 7 }\] \[\frac{ 9^x }{ 7^x }=\frac{ 1 }{ 7 }\] \[\left( \frac{ 9 }{ 7 } \right)^x=\frac{ 1 }{ 7 }\] take log and find x
anonymous
  • anonymous
−7.74293 7.74293 −1 1 These are the answer choices <3
anonymous
  • anonymous
Not quite. Let's look at just the left hand side. You are trying to determine\[\log 3^{2x} = ?\]Compare that with the general case\[\log a^b = b \log a\]Substitute a = 3 and b = 2x. What do you get?
anonymous
  • anonymous
2x log 3
anonymous
  • anonymous
was i right
anonymous
  • anonymous
Correct. Now do the same thing on the right hand side. You are trying to determine\[\log 7^{x-1} = ?\]Applying the same rule, what do you get?
anonymous
  • anonymous
x-1=log 7
anonymous
  • anonymous
yes?
anonymous
  • anonymous
Not exactly. In this one, a = 7 and b = x-1. Try again
anonymous
  • anonymous
7 log x-1
anonymous
  • anonymous
Backwards. Try again
anonymous
  • anonymous
x-1 log 7??
anonymous
  • anonymous
i got -7.74293 as an answer
anonymous
  • anonymous
That's it. So now you done\[\log 3^{2x} = \log 7^{x-1}\]\[2x \log 3 = \left( x-1 \right) \log 7\]OK so far?
anonymous
  • anonymous
yes! I am good
anonymous
  • anonymous
Great. Now expand the right hand side.
anonymous
  • anonymous
0.47712125472(2x)
anonymous
  • anonymous
That's the left hand side. You can multiply the 0.477... by the 2.
anonymous
  • anonymous
0.95424250943 sorry
anonymous
  • anonymous
Excellent. The left hand side is 0.95424250944 x. Now, on to the left hand side. First thing to do is to expand it.
anonymous
  • anonymous
** right hand side. Sorry
anonymous
  • anonymous
0.84509804001x-0.845098804001
anonymous
  • anonymous
right??
anonymous
  • anonymous
Exactly well done! So now you have\[0.95424250944 x = 0.84509804001 x - 0.8450904001\]Can you gather up the x's on one side and solve?
anonymous
  • anonymous
-0.36797678529=-0.8450904001
anonymous
  • anonymous
so would the answer be 7.74293?
anonymous
  • anonymous
Problem with the left hand side. Remember, you are subtracting 0.84509804001 x from both sides. Try the left hand side again.
anonymous
  • anonymous
0.10914370543x
anonymous
  • anonymous
That's better. So you have\[0.10914370543 x = 0.84509804001\]To solve for x, divide both sides by 0.10914370543. What do you get?
anonymous
  • anonymous
7.74298468868 !!!!
anonymous
  • anonymous
Yayyyy!! Well done!
anonymous
  • anonymous
Yayyyy! If I post another question in the open section, will u answer?
anonymous
  • anonymous
thanks:)
anonymous
  • anonymous
Yup. You're welcome
anonymous
  • anonymous
To avoid dealing with all those decimal places, most folks will leave the logs until the end, for example\[2x \log 3 = (x-1)\log 7\]\[(2 \log 3) x = (\log 7) x - \log 7\]\[(2 \log 3 - \log 7) x = -\log 7\]\[x = \frac{ -\log 7 }{ 2 \log 3 - \log 7 }\]Then plug it into your calculator

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