alg2 logarithms?

Question

Please help me please. I need a clear step-by-step explanation on a few problems that I'm stuck on. Thank you!!!!

1) Solve for x in terms of a and b: (a^2 – 2a^2b^2 + b^4)^(x-1) = (a - b)^(2x)(a + b)^(-2)


2) Given that 5^ [log (base 5)][(log (base2) log(base3) log (base4) x = log (base squareoot5 – 2) (9 – 4squareroot5) and 1/i(as in imaginary i)log^[log(base e)log(base2)y] = pi. Evaluate log(base 2) (x/y). (HINT: e^(pi (imaginary i)) = -1


3) Solve for x: (5 squareroot2 x) ^ [(log (base4)) (5 squareroot2 x)] = 625 x ^ 4


4) Solve for x: log(base8)(1 - x) + (10log(base32)x) / (log(base8) 512) – log (base2)e^[ln(1/x+1)] / 243^(1/3) = cuberoot(64/27)

Answer 1

1) Solve for x in terms of a and b: (a^2 – 2a^2b^2 + b^4)^(x-1) = (a - b)^(2x)(a + b)^(-2)
use log:
log (a^2 – 2a^2b^2 + b^4)^(x-1) = log ((a - b)^(2x)(a + b)^(-2) )
(x-1)log (a^2 – 2a^2b^2 + b^4) = 2x log(a-b) + log(a+b)^{-2}
(x-1) log ( (a-b^2)^2 ) = 2x log(a-b) -2log(a+b)
2x log(a-b^2) -2x log(a-b) = -2log(a+b) + 2log(a-b)
2x( log(a-b^2) -log(a-b) ) = -2log(a+b) + 2log(a-b)
x= ( -2log(a+b) + 2log(a-b)) /2( log(a-b^2) -log(a-b) )
x= ( -log(a+b) + log(a-b)) /( log(a-b^2) -log(a-b) ).

Answer 2

Man that is a lot of work. I can help with with 1.

(a^2 – 2a^2b^2 + b^4)^(x-1) = (a - b)^(2x)(a + b)^(-2)

Start by taking the log of both sides

log (a^2 – 2a^2b^2 + b^4)^(x-1) = log (a - b)^(2x)(a + b)^(-2)

Remember the log of a product is the sum of the logs.

(x-1) log (a^2 – 2a^2b^2 + b^4) = log (a-b) ^2x + log (a+b)^(-2)

-->This is where you are getting confused on the left side:

(x-1) * log (a^2 – 2a^2b^2 + b^4)
x * log (a^2 – 2a^2b^2 + b^4) - log (a^2 – 2a^2b^2 + b^4) = log (a-b) ^2x + log (a+b)^(-2)




(x-1) log (a^2 – 2a^2b^2 + b^4) = 2x log (a-b) - 2 * log (a+b)

Group the x and constants to one side

x*log (a^2 – 2a^2b^2 + b^4) - 2xlog (a-b) = log (a^2 – 2a^2b^2 + b^4) - 2 * log (a+b)

All you have to do is factor the x on left and divide both side and solve for x.

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