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

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 $\log_a(1)=0$

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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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.

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

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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