Asked by Bob

Determine a value of 𝑥 that makes the following statement true.
𝑙𝑜𝑔2(3)𝑥 + 𝑙𝑜𝑔2 (x/81(y^2-x^2)) + 𝑙𝑜𝑔(𝑥)/log(2) = 𝑙𝑜𝑔2(𝑥^2) + 𝑙𝑜𝑔2(𝑦 − 𝑥)^−1 − 𝑙𝑜𝑔2(𝑥 + 𝑦)
Answered by oobleck
not sure just what log2(3)x means, but if it log2(3)*x, then
log2(x/81 (y^2-x^2)) = log2x - log2(81) + log2(y^2-x^2)
log2(x/(81 (y^2-x^2)) = log2x - log2(81) - log2(y^2-x^2)
logx/log2 = log2(x)
So, since all the logs are nase 2, we can just say that

(log3 x)(x/(81(y^2-x^2)))(x) = (x^2)/((y-x)(y+x))
(log3 x) = x/81

That's as far as I can go without knowing just what log2(3)x is supposed to mean. Maybe you can finish it off yourself now.
Answered by Bob
Hey there! Really sorry with the confusion on log2(3)x.
x is supposed to be an exponent but I completely forgot to put the ^x.
𝑙𝑜𝑔2(3)^𝑥 + 𝑙𝑜𝑔2 (x/81(y^2-x^2)) + 𝑙𝑜𝑔(𝑥)/log(2) = 𝑙𝑜𝑔2(𝑥^2) + 𝑙𝑜𝑔2(𝑦 − 𝑥)^−1 − 𝑙𝑜𝑔2(𝑥 + 𝑦).

Answered by oobleck
still kinda bogus, since what you have written really means
log2(3)^x = (log2(3))^x
when what you wanted was log2(3^x)
ok, so now we're stuck with

log_2(3^x) = x log3, so
x log3 = x/81
81 log3 = 1
bzzzt - but thanks for playing
and x is completely gone from the picture. So check my interpretations of things, and maybe you will get to the point where
x = log_3(81) = 4
There are no AI answers yet. The ability to request AI answers is coming soon!