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)

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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)

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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.
@whpalmer4 sorry its \[125^{9x - 2} = 150\]
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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You would still do the same thing. Take the log of both sides, and solve for x.
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.
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
\[\log(x+9) - \log x = 3\]\[\log(\frac{x+9}{x}) = 3\]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
Ok I don't know how you got that but that would mean x+9/x = 20.086 right?
Yes, e^3 = 20.086 (approximately)
But you can put it in a form x =
Multiply both sides by x, then subtract x from both sides
and divide by e^3
so x+9 = 20.086x 9 = 20.086x - x 0.448 = 0?
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
Did that make sense?
So 0.472?
If it helps, these are the multiple choice answers for the problem. 0.0090 0.3103 3.2222 111
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.
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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