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Mathematics 21 Online
OpenStudy (anonymous):

Use the One-to-One Property to solve the equation for x. (Picture below)

OpenStudy (anonymous):

OpenStudy (swissgirl):

what is the one to one property? Like this? We know that \(2^7=128 \) So \(2^{3x}=128 \to 2^{3x}=2^7\) so \(3x=7\) Hence \(x=\frac{7}{3}\)

OpenStudy (anonymous):

x= 7/3

OpenStudy (swissgirl):

Not sure if this was the method u were looking for

OpenStudy (anonymous):

let me get a screen shot of the one-to-one property. hold on one second.

OpenStudy (anonymous):

OpenStudy (swissgirl):

ok so I used the correct method. I figured it was this. So basically the one to one method states that is the base is the same and the right hand value equals the left hand value then the exponents must equal each other

OpenStudy (anonymous):

ohh. that makes sense. the teacher only went over one of these, we ran out of time. can you help me with this last one?

OpenStudy (swissgirl):

Ya no problem

OpenStudy (anonymous):

OpenStudy (swissgirl):

Ok we need to use log properties for this one. Are u familiar with them?

OpenStudy (anonymous):

like changing them to simplify basically? like adding, subtracting, etc?

OpenStudy (swissgirl):

Log Rules: 1) \( log_b(mn) = log_b(m) + log_b(n)\)

OpenStudy (anonymous):

okay, yes i know those.

OpenStudy (swissgirl):

ok so \( log_3(x)+log_3(x-2)=log_3(x*(x-2))\) So \( log_3(x)+log_3(x-2)=1 \to log_3(x*(x-2))=1\) Can you continue from here?

OpenStudy (anonymous):

uhh. no i can't lol. i remember doing one like this.. but not exactly.

OpenStudy (swissgirl):

ok give me a few seconds

OpenStudy (anonymous):

okay

OpenStudy (swissgirl):

ok back

OpenStudy (swissgirl):

so \(log_3(x*(x-2))=1 \to 3^1=x*(x-2)\) \(3=x^2-2x \to 0=x^2-2x-3\) \(0=(x-3)(x+1)\) Hence x=3 and -1

OpenStudy (swissgirl):

I am assuming you are familiar with quadratics and log properties but ask if you dont follow any step

OpenStudy (anonymous):

ohh! no that makes sense! thanks so much!

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