請問以下呢條數點做? Solve 8^(x-2)=3(4^2x)?

設 t=2^x
8^(x-2)=3(4^2x)
2^(3x-6)=3*2^(4x)
t^3/64=3t^4
t^3/192=t^4
t^3(1-192t)=0
t=0 or 1/192
2^x=192
x=log (192,2)=6+log(3,2)

8^(x-2) = 3(4^2x)
2^3(x-2) = 3(2^2)^(2x)
2^(3x-6) = 3(2^4x)
(2^3x) [2^(-6)] = 3(2^4x)
2^(-6) = 3(2^x)
log 3(2^x) = log 2^(-6)
log 3 + x log 2 = -6 log 2 - - - - - - -《Formula: log ab = log a + log b ;　log aˣ = x log a 》
(x+6) log 2 = - log 3
∴ x = - (log 3 / log 2) - 6

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 完 ~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

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可用 「代入」的方法 check 答案 !

計出 x = - (log 3 / log 2) - 6 ≒ -7.58
Putting x= -7.6 into 8^(x-2) and then into 3(4^2x), both come up to around the same answer.