if 2^x+1=3^1-x then find the value of x
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Take log on both sides
(x+1)log2=(1-x)log3
Alternatively, (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
x= 0.2263
Verification:
LHS=2^(1.2263)=2.3397
RHS=3^(0.7737)=2.3397
Hence proved!
hope it helps
(x+1)log2=(1-x)log3
Alternatively, (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
x= 0.2263
Verification:
LHS=2^(1.2263)=2.3397
RHS=3^(0.7737)=2.3397
Hence proved!
hope it helps
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