Question

pls help me
${a}^{3 - x} {b}^{5x} = {a}^{x + 5} {b}^{3x} \\ then \: x log(b \div a) =$
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Answer 1

Answer:

a^(3-x) * b^(5x)=a^(5+x) * b^(3x)

a^(3-x) * b^(5x)

---------------------- = 1

a^(5+x) * b^(3x)

a^[(3-x) - (5+x)] * b^(5x - 3x) = 1

a^[3 - x - 5 - x] * b^(2x) = 1

a^(-2x - 2) * b^(2x) = 1

a^[-2(x+1)] * b^(2x) = 1

. .b^(2x)

---------------- = 1

a^[2(x+1)]

b^(2x) = a^[2(x+1)] ---> we log both sides now

log{ b^(2x) } = log{ a^[2(x+1)] }

2x*log(b) = 2(x+1)*log(a). . . |:2

x*log(b) = (x+1)*log(a)

x*log(b) = x*log(a) + log(a)

x*log(b) - x*log(a) = log(a)

x*[ log(b) - log(a) ] = log(a)

x*log(b/a) = log(a) --->QED

Hope it helps, and if there is a step you didn't get tell me!

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