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logarithms- alg 2
I can't figure out some of these log problems. I asked my teacher for help yesterday and she gave me some quick, vague information and i still don't how to do some of them. Oh, and don't tell me to go read my text book, because we don't have one. Does anyone know how to do these?
log9(x + 2)= 0 X=?
oh, and the 9 is actually base 9.
logbase6(x+9) + logbase6 x=2 x=?
anyone know??
thanks.
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logbase9(x+2)= 0
9^0=x+2**any number raised to 0 is equal to 1
so, 1=x+2
X=-1
logbase6(x+9)+logbase6x =2
they both have the same base of logbase6, and there being added, so theres a rule that u multiply them
longbase6(x+9)(x)=2
X^2+9x=2
X^2+9x-2=0
factor that
logs suck.
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i think the first one goes like this
9^0=x+2 nine to the power of 0 equals x plus 2
1=x+2 nine to the power of 0 is one
x=-1 one minus two equals negative one
i dont remember how to do the second one, but im pretty sure the first one is right.
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