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\[\log_{3x+7}(5x+3)+\log_{5x+3}(3x+7)=2 \]
Do you know the log rules?
some of them
which one do I need
Do you know the change of base rule? The multiplication/division rules?
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yesssss
Use the change of base rule to a common base such as e, for BOTH of the logs. Then you will find you can cancel out the fractions
I don't see how they cancel
Oh I thought you were supposed to prove it
no solve
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Find a value of x that makes 5x+3 = 3x+7. Then, you get 1+1=2.
Because \(\log_{b}b = 1\)
@ArkGoLucky, does that make sense?
That makes sense but I don't know how to do that
Do you mean you can't solve for x? Or that you can't see why we need to do that.
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nevermind I understand. thanks
Okay, great!
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