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.
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)
3 answers
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).
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).
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
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