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Solve the equation of 2log (2y + 4) = log 9 + 2log (y -1).
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A. 3 B. 5 C. 7 D. 21
How would i solve this ?
Condense the logs on the right side first.
log9+log2 first or log2(x-1)?
first bring the exponents up like this \[\Large \log (2y + 4)^2 = \log 9 + \log (y -1)^2\] then condense logs on the right
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Sorry not sure how to do this could you walk me throught it?
Use the log rule log(a) + log(b) = log(a)(b)
eg log(x+3) + log(x-2) = log(x+3)(x-2)
log(9)(2y-1)^2
\[\Large \log (2y + 4)^2 = \log 9(y -1)^2 \]now you can just drop the logs on both sides, and solve for y. Logs are one-to-one functions so you can do that here. eg if logx = logy then x = y
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(2y+4)^2=9(2y-1)^2 would be the qeuation to solve
Yep
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