solve the equation log base 3 (9x) = log base x (27)
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log(base 3) 9x = log(base x) 27
ln(9x)/ln(3) = ln(27)/ln(x) => (ln(9) + ln(x))/ln(3) = ln(27)/ln(x)
ln(x)ln(9) + ln²(x) = ln(27)ln(3)
Let z = ln(x)
z² + ln(9)z - ln(27)ln(3) = 0
z² + 2ln(3)z - 3ln²(3) = 0
(z + 3ln(3))(z - ln(3)) = 0
ln(x) = -3ln(3) => x = 1/27
ln(x) = ln(3) => x = 3
ln(9x)/ln(3) = ln(27)/ln(x) => (ln(9) + ln(x))/ln(3) = ln(27)/ln(x)
ln(x)ln(9) + ln²(x) = ln(27)ln(3)
Let z = ln(x)
z² + ln(9)z - ln(27)ln(3) = 0
z² + 2ln(3)z - 3ln²(3) = 0
(z + 3ln(3))(z - ln(3)) = 0
ln(x) = -3ln(3) => x = 1/27
ln(x) = ln(3) => x = 3
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change both to base 10
log(9x) /log(3) = log(27)/log(x)
The two sides are equal when x= 3
log(9x) /log(3) = log(27)/log(x)
The two sides are equal when x= 3
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log base 3 9x = log base x 27
log base b a /log b base c = log base c b
log 9x/log 3 = log 27/log x
9x/3 = 27/x
9x^2 = 81
x^2 = 9
x = 3
log base b a /log b base c = log base c b
log 9x/log 3 = log 27/log x
9x/3 = 27/x
9x^2 = 81
x^2 = 9
x = 3