rosedewittbukater
  • rosedewittbukater
Algebra 2 help please? And explanation?? Graph the logarithmic equation. y = log5(x – 2) Solve the exponential equation. 1259x – 2 = 150 Solve the equation. log(x + 9) – log x = 3 Write the expression as a single natural logarithm. 3 ln a - 1/2(ln b + ln c^2)
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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whpalmer4
  • whpalmer4
Solve the exponential equation - the equation doesn't make much sense. Can you rewrite it using the Equation button? Is it \[1259^{x-2} = 150\]If so, take the log of both sides and solve for x. To solve \[\log(x+9) - \log x = 3\] remember that subtracting two logarithms is like dividing the two numbers and then taking the log. For example, \[\log 6-\log 3 = \log 2\] because 6/3 = 2. After you've converted the left hand into the log of a fraction, do e^x on both sides to get rid of the log and solve for x. For the last one, remember that just as subtracting two logarithms is like dividing and taking the log, adding two logarithms is like multiplying and taking the log. See what you can do with those hints.
rosedewittbukater
  • rosedewittbukater
@whpalmer4 sorry its \[125^{9x - 2} = 150\]
rosedewittbukater
  • rosedewittbukater
For the second one I was able to get to \[\log_{\frac{ x + 9 }{ 9 }} = 3\] but I don't know what to do next.

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whpalmer4
  • whpalmer4
You would still do the same thing. Take the log of both sides, and solve for x.
rosedewittbukater
  • rosedewittbukater
For the third one I got to \[3\ln a - \frac{ 1 }{ 2 } (\ln b c ^{2})\] But I'm still stuck. I'm stuck on the other two as well.
rosedewittbukater
  • rosedewittbukater
How can I solve he first one if the x is in the exponent? And for the second one I meant log x+9/x
whpalmer4
  • whpalmer4
\[\log(x+9) - \log x = 3\]\[\log(\frac{x+9}{x}) = 3\]Raise e^ both sides \[e^{\log(\frac{x+9}{x})} = e^3\]but\[ e^{\log x} = x\] so that simplifies to \[\frac{x+9}{x} = e^3\] And I think you can solve that
rosedewittbukater
  • rosedewittbukater
Ok I don't know how you got that but that would mean x+9/x = 20.086 right?
whpalmer4
  • whpalmer4
Yes, e^3 = 20.086 (approximately)
whpalmer4
  • whpalmer4
But you can put it in a form x =
whpalmer4
  • whpalmer4
Multiply both sides by x, then subtract x from both sides
whpalmer4
  • whpalmer4
and divide by e^3
rosedewittbukater
  • rosedewittbukater
so x+9 = 20.086x 9 = 20.086x - x 0.448 = 0?
whpalmer4
  • whpalmer4
Oh, stick with the symbolic form until the bitter end: x+9 = e^3 x 9 = e^3 x - x 9 = x(e^3 - 1) x = 9 / (e^3 -1) now get out the calculator if you need a numeric representation
whpalmer4
  • whpalmer4
Did that make sense?
rosedewittbukater
  • rosedewittbukater
So 0.472?
rosedewittbukater
  • rosedewittbukater
If it helps, these are the multiple choice answers for the problem. 0.0090 0.3103 3.2222 111
whpalmer4
  • whpalmer4
Okay, the problem is using log base 10 in that case. So instead of e^3, you have 10^3 = 1000 and you'll find the answer in the list.
whpalmer4
  • whpalmer4
For this one: \[3 \ln a - \frac{1}{2}(\ln b + \ln c^2)\] adding logs of two numbers is like taking the log of the product, so \[3 \ln a - \frac{1}{2}(\ln b + \ln c^2) = 3 \ln a - \frac{1}{2}\ln (bc^2)\] Also, multiplying the log of x by a number is like raising x to that number, then taking the log. So we can convert that 3 ln a to ln a^3 and similarly 1/2 ln (bc^2) becomes ln (sqrt(bc^2)) = ln (c sqrt(b)) because raising something to the 1/2 power is taking the square root (in general, raising to 1/n is taking the nth root) Then after you've done those simplifications, remember that subtracting logs is like taking the log of the first number divided by the second.

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