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secondary 3 | A Maths
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i dont know how to start
log9(y²) = 3 + logy(81)
Theres a log9(y)² on the left so you have to change the right side's terms into that form.
2log9(y) = 3 + logy(9²)
2log9(y) = 3 + 2logy(9)
2log9(y) = 3 + 2 (1/log9(y) )
2log9(y) = 3 + 2/log9(y)
Multiply by log9(y) to get rid of the denominator,
2(log9(y))² = 3log9(y) + 2
Now you can see that the equation is in the form of 2u² = 3u + 2
log9(y²) = 3 + logy(81)
2log9(y) = 3 + logy(9²)
2(1/logy(9) ) = 3 + 2logy(9)
Multiply by logy(9),
2 = 3logy(9) + 2( logy(9) )²
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