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9^x-1=27
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\[2(x-1)=3\] is a start
9^x = 28 x ln 9 = ln 28 x = ln 28/ln 9
9^x-1=27 9^x=28 log9^x=log28 xlog9=log28 x=log28/log9 x=1.516551628
27 = 3^3 9 = 3^2 so (3^2)^x-1 = 3^3 3^2x-2 =3^3 since base is same, powers have to be equal so 2x - 2 = 3 2x = 3+2 x = 5/2
To three significant digits, x = 1.52. If the original problem is supposed to be 9^(x - 1) = 27, then Harkirat is correct.
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Those parentheses make a big difference.
true!! without them it is difficult to decide which way to go i went one way and you the other.........
Well, at least all bases are covered:^)
i am willing to bet based on the other problems that it is \[9^{x-1}=27\]
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