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Solve exactly. log(base 6)(x+1) + log(base 6)(x+2) = 1 x= ?
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I though it would work out to (x+1) = 1-(x+2)
based on what you had told me earlier
no this is a different situation.
You need to use properties of logarithms in this one.
no first combine into 1 log
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When you have a sum of logs, you can make it into the log of a product: \[log_ba + log_bc = log_b(a\cdot c)\]
log_6(x+1)(x+2) = 1 (x+1)(x+2) = 6 ...
x^2+3x+2=6
Right, then solve the quadratic.
subtract 6 and equal to 0?
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Yep
x^2+3x-4=0 so (x+4)(x-1)
perfect
and a log cannot be negative correct? so x=1 ?
yes
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awesome you two have been huge helps
np
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