Solving equations with logs: log (2x + 5) * log (9x^2) = 0

Quadratic equations and inequalities, variation equations, function notation, systems of equations, etc.
santaclaus
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Solving equations with logs: log (2x + 5) * log (9x^2) = 0

Postby santaclaus » Fri Apr 10, 2009 9:59 pm

how do I go about solving?

log (2x+5) x log (9x2)=0

I would start by distributing log to 2x and then log 5? and then go from there?

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Martingale
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Re: Solving equations with logs

Postby Martingale » Fri Apr 10, 2009 11:07 pm

santaclaus wrote:how do I go about solving?

log (2x+5) x log (9x2)=0

I would start by distributing log to 2x and then log 5? and then go from there?


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santaclaus
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Re: Solving equations with logs: log (2x + 5) * log (9x^2) = 0

Postby santaclaus » Mon Apr 13, 2009 6:58 am

how do you know that log 2x+5=1? where you wrote the answer? Is it true that log A * log B = log A + log B =0? I'm not sure where you're getting 2 equations =1?

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stapel_eliz
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Postby stapel_eliz » Mon Apr 13, 2009 11:01 am

santaclaus wrote:how do you know that log 2x+5=1?

You don't. You know (by factoring, like you learned back when you were studying quadratic equations), that log(2x + 5) = 0. By using what you know about logs, you then converted this log equation into its equivalent exponential equation. :wink:

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Martingale
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Re: Solving equations with logs: log (2x + 5) * log (9x^2) = 0

Postby Martingale » Mon Apr 13, 2009 10:31 pm

santaclaus wrote:how do you know that log 2x+5=1? where you wrote the answer? Is it true that log A * log B = log A + log B =0? I'm not sure where you're getting 2 equations =1?




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