I need help on this question.
9^x + 7(3^x-1) = 16
Thanks in advance.
9^x + 7(3^x-1) = 16
Thanks in advance.
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9^x + 7(3^x-1) = 16
3^(2x) + 7(3^x) - 23 = 0
Let Y = 3^x,
Y^2 + 7Y - 23 = 0
Y = (-7 ± √(49 + 92))/2
Y = (-7 ± 11.874)/2
Y = 2.437 or Y = -9.437 (discard)
When Y=2.437,
3^x=2.437
x log3 = log 2.437
x = (log 2.437)/(log 3) = 0.824
3^(2x) + 7(3^x) - 23 = 0
Let Y = 3^x,
Y^2 + 7Y - 23 = 0
Y = (-7 ± √(49 + 92))/2
Y = (-7 ± 11.874)/2
Y = 2.437 or Y = -9.437 (discard)
When Y=2.437,
3^x=2.437
x log3 = log 2.437
x = (log 2.437)/(log 3) = 0.824