if 2^(x+1)=3^(1-x), then find the value of x
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HEYA!!!
HERE IS YOUR ANSWER,
=> Take log on both sides,
=> (x+1)log2 = (1-x)log3
=> (x+1)/(1-x) = log3/log2
=> Take componendo and dividendo on both sides:
=> 2/2x = (log3+log2)/(log3-log2)
=> Take reciprocal,
=> x = (log3-log2)/(log3+log2)
=> Since the values for Log3 = 0.4771 and Log2 = 0.3010
Therefore,
=> x = 0.2263
HOPE IT HELPS YOU,
THANK YOU.☺️☺️
HERE IS YOUR ANSWER,
=> Take log on both sides,
=> (x+1)log2 = (1-x)log3
=> (x+1)/(1-x) = log3/log2
=> Take componendo and dividendo on both sides:
=> 2/2x = (log3+log2)/(log3-log2)
=> Take reciprocal,
=> x = (log3-log2)/(log3+log2)
=> Since the values for Log3 = 0.4771 and Log2 = 0.3010
Therefore,
=> x = 0.2263
HOPE IT HELPS YOU,
THANK YOU.☺️☺️
YagamiLight:
what is componendo and dividendo
See, if you have a/b and if you apply componendo and dividendo then your answer would be (a+b)/(a-b)
Answered by
3
math]2^{x+1}=3^{1-x}[/math]
[math](e^{\ln(2)})^{x+1}=(e^{\ln(3)})^{1-x}[/math]
[math]e^{\ln(2)(x+1)}=e^{\ln(3)(1-x)}[/math]
[math]\ln(2)(x+1)=\ln(3)(1-x)[/math]
[math]\ln(2)x+\ln(2)=\ln(3)-\ln(3)x[/math]
[math](\ln(2)+\ln(3))x=\ln(3)-\ln(2)[/math]
[math]\ln(6)x=\ln(3)-\ln(2)[/math]
[math]x=\frac{\ln(3)-\ln(2)}{\ln(6)}[/math]
[math](e^{\ln(2)})^{x+1}=(e^{\ln(3)})^{1-x}[/math]
[math]e^{\ln(2)(x+1)}=e^{\ln(3)(1-x)}[/math]
[math]\ln(2)(x+1)=\ln(3)(1-x)[/math]
[math]\ln(2)x+\ln(2)=\ln(3)-\ln(3)x[/math]
[math](\ln(2)+\ln(3))x=\ln(3)-\ln(2)[/math]
[math]\ln(6)x=\ln(3)-\ln(2)[/math]
[math]x=\frac{\ln(3)-\ln(2)}{\ln(6)}[/math]
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