## 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

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?

Martingale
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### Re: Solving equations with logs

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 $\log_a(1)=0$

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

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?

stapel_eliz
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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.

Martingale
Posts: 333
Joined: Mon Mar 30, 2009 1:30 pm
Location: USA
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### Re: Solving equations with logs: log (2x + 5) * log (9x^2) = 0

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