Asked by anju
solve 3^(2x-3)=5*2^(x^2) for x
Answers
Answered by
Reiny
take log of both sides and use your log rules
log (3^(2x-3)) = log [5*2^(x^2))
(2x-3)(log3) = log5 + (log2)x^2
2log3 x - 3log3 = log5 + log2(x^2)
log2(x^2) - 2log3 x + log5 + 3log3 = 0
log2(x^2) - 2log3 x + log 135 = 0
this is quadratic with
a = log2 , b = 2log3 or log9 , and c = log 135
x = (-log9 ± √((log9)^2 - 4(log2)(log135)))/2log2
carefully work it out on your calculator, I got
the discriminant to be negative, so , after all that, there is no real solution
Making a rough sketch of y = 3^(2x-3) and y = 5*2^(x^2) will show that they don't meet.
log (3^(2x-3)) = log [5*2^(x^2))
(2x-3)(log3) = log5 + (log2)x^2
2log3 x - 3log3 = log5 + log2(x^2)
log2(x^2) - 2log3 x + log5 + 3log3 = 0
log2(x^2) - 2log3 x + log 135 = 0
this is quadratic with
a = log2 , b = 2log3 or log9 , and c = log 135
x = (-log9 ± √((log9)^2 - 4(log2)(log135)))/2log2
carefully work it out on your calculator, I got
the discriminant to be negative, so , after all that, there is no real solution
Making a rough sketch of y = 3^(2x-3) and y = 5*2^(x^2) will show that they don't meet.
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