solve, giving your answers as exact fractions, the simultaneous equations: 8^(x) = 4^(2x+3) log(base2)y = log(base2)x + 4 ('^' meaning 'to the power of') I am stuck, i dont know how to do it! PLease show your method, Thanks, azza xxxz
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Originally Posted by azza solve, giving your answers as exact fractions, the simultaneous equations: 8^(x) = 4^{2x+3} is the same as . Equate the exponents.
Originally Posted by azza solve, giving your answers as exact fractions, the simultaneous equations: 8^(x) = 4^(2x+3) log(base2)y = log(base2)x + 4 I assume the right side of the second equation is rather than ... can you finish it?
sorry i typed it in incorrecty 8^(y) = 4^(2x+3) log(base2)y = log(base2)x + 4 so that gives you 3y = 4x +6 which i make into y = 4/3x + 2 and I put it into the y of the other equation. but how do i solve it then?
y = 16x ... substitute 16x for y into the first equation.
thankyou very much! i am getting x=1/8, y=2, but the textbook answer is x=3/22 y=24/11 is the textbook wrong or am i?
Originally Posted by azza thankyou very much! i am getting x=1/8, y=2, but the textbook answer is x=3/22 y=24/11 is the textbook wrong or am i? the textbook is correct ... recheck your algebra.
ahh finally i understand, thankyou so much!
I have ONe other question: log(base 3)(2-3x) = log(base9)(6x^2 - 19x + 2) Thankyou x
Originally Posted by azza I have ONe other question: log(base 3)(2-3x) = log(base9)(6x^2 - 19x + 2) Thankyou x note that next time start a new problem w/ a new thread.
sorry, i dont see where that leads?
finish it
Another way of looking at it: [tex]log_a(x)= y if and only if so if then 2- 3x= 3^y[tex] and if then . But so . Solve
Originally Posted by HallsofIvy Another way of looking at it: [tex]log_a(x)= y if and only if so if then 2- 3x= 3^y[tex] and if then . But so . Solve the last equation should be ...
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