SOLUTION: Solve the logarithmic equation algebraically. Round to three decimal places. log base(4)5x- log(4)(6+sq root of x)=2

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Question 369496: Solve the logarithmic equation algebraically. Round to three decimal places.
log base(4)5x- log(4)(6+sq root of x)=2
Answer by CharlesG2(834) About Me  (Show Source):
You can put this solution on YOUR website!
Solve the logarithmic equation algebraically. Round to three decimal places.
log base(4)5x- log(4)(6+sq root of x)=2

log4 (5x) - log4 (6 + sqrt(x)) = 2
logarithmic rule: logb (m/n) = logb (m) - logb (n)
log4 (5x/(6 + sqrt(x))) = 2
logarithmic rule: logb (m) = logk (m)/ logk b, where k is your new base
log4 (5x/(6 + sqrt(x))) = [log10 (5x/(6 + sqrt(x)))] / log10 (4) = 2
log10 (5x/(6 + sqrt(x))) = 2log10 (4)
logarithmic rule: nlogb (m) = logb (m^n)
4^2 = 16
log10 (5x/(6 + sqrt(x))) = log10 (16)

5x/(6 + sqrt(x)) = 16
5x = 16(6 + sqrt(x))
5x = 96 + 16sqrt(x)
5x - 96 = 16sqrt(x)
(5x - 96)^2 = (16sqrt(x))^2 (square both sides)
25x^2 - 960x + 9216 = 256x
25x^2 - 1216x + 9216 = 0, solve this and determine which root works

use quadratic formula, a = 25, b = -1216, c = 9216
x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+
x+=+%281216+%2B-+sqrt%281478656+-+4%2A25%2A9216%29%29%2F50+
x+=+%281216+%2B-+sqrt%281478656+-+921600%29%29%2F50+
x+=+%281216+%2B-+sqrt%281478656+-+921600%29%29%2F50+
x+=+%281216+%2B-+sqrt%28557056%29%29%2F50+
557056 = 128 * 128 * 34
x+=+%281216+%2B-+128sqrt%2834%29%29%2F50+
x+=+%28608+%2B-+64sqrt%2834%29%29%2F25+

x1 = 608/25 + (64/25)sqrt(34)
x1 = 24.32 + 2.56sqrt(34)
x1 = 39.247237 to 6 places = 39.247 to 3 places
x2 = 24.32 - 2.56sqrt(34)
x2 = 9.392763 to 6 places = 9.392 to 3 places

check log10 (5x/(6 + sqrt(x))) = log10 (16)
log10 (16) = 1.204120 to 6 places
x1 gives log10 (16) = log10 (16), x1 works
x2 gives log10 (5.180921) which does not equal log10 (16), x2 does not work

answer for x is 39.247 to 3 places