Asked by YOY
I need help with a question!
If 3^(3x+2)*7^(x-1)=81^x*7^2x,
what is 21^x
If 3^(3x+2)*7^(x-1)=81^x*7^2x,
what is 21^x
Answers
Answered by
Henry
3^(3x+2) * 7^X-1) = (3^4)^x * 7^2x.
3^(3x+2) * 7^(x-1) = 3^4x * 7^2x.
Divide both sides by 3^4x*7^2x:
3^(3x+2)*7^(x-1) / (3^4x*7^2x) = 1.
Divide the denominator and numerator by
3^4x*7^2x:
3^(-x+2)*7^(-x-1) = 1.
(-x+2)*Log3 + (-x-1)*Log7 = Log1.
(-x+2)0.477 + (-x-1)0.845 = 0.
-0.477x + 0.954 - 0.845x -0.845 = 0.
-1.322x + 0.109 = 0.
1.322x = 0.109.
X = 0.109 / 1.322 = 0.08245.
3^(3x+2) * 7^(x-1) = 3^4x * 7^2x.
Divide both sides by 3^4x*7^2x:
3^(3x+2)*7^(x-1) / (3^4x*7^2x) = 1.
Divide the denominator and numerator by
3^4x*7^2x:
3^(-x+2)*7^(-x-1) = 1.
(-x+2)*Log3 + (-x-1)*Log7 = Log1.
(-x+2)0.477 + (-x-1)0.845 = 0.
-0.477x + 0.954 - 0.845x -0.845 = 0.
-1.322x + 0.109 = 0.
1.322x = 0.109.
X = 0.109 / 1.322 = 0.08245.
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