The Purplemath Forums

by **santaclaus** on 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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by **Martingale** on 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?

if the problem is $\log_a(2x+5)\cdot\log_a(9x^2)=0$

then you need

$\log_a(2x+5)=0$ or $\log_a(9x^2)=0$

so $2x+5=1$ or $9x^2=1$

since $\log_a(1)=0$

by **santaclaus** on 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?

by **stapel_eliz** on 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:

by **Martingale** on 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?

$\log_a(1)=0$

so $\log_a(2x+5)=0 \Leftrightarrow 2x+5=1$

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

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