SOLUTION: solve algebraically: log base 2(3x+2) - log base 4(x)=3

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Question 118903: solve algebraically:
log base 2(3x+2) - log base 4(x)=3
Found 2 solutions by stanbon, jim_thompson5910:
Answer by stanbon(48569) About Me  (Show Source):
You can put this solution on YOUR website!
solve algebraically:
log base 2(3x+2) - log base 4(x)=3
------------
log (base 2)(3x+2) - log(base 2)(2x) = 3
log[(3x+2)/2x] = 3
(3x+2)/2s = 2^3
3x+2 = 16x
12x = 2
x = 1/6
==============
Cheers,
Stan H.

Answer by jim_thompson5910(21685) About Me  (Show Source):
You can put this solution on YOUR website!
log%282%2C%283x%2B2%29%29-log%284%2C%28x%29%29=3 Start with the given equation



log%2810%2C%283x%2B2%29%29%2Flog%2810%2C%282%29%29-log%2810%2C%28x%29%29%2Flog%2810%2C%284%29%29=3 Use the change of base formula log%28b%2Cx%29=log%2810%2Cx%29%2Flog%2810%2Cb%29 to rewrite every log in base 10


log%2810%2C%283x%2B2%29%29%2Flog%2810%2C%282%29%29-log%2810%2C%28x%29%29%2Flog%2810%2C%284%29%29=3 Rewrite log%2810%2C%284%29%29 as log%2810%2C2%5E2%29


log%2810%2C%283x%2B2%29%29%2Flog%2810%2C%282%29%29-log%2810%2C%28x%29%29%2F2log%2810%2C%282%29%29=3 Rewrite log%2810%2C2%5E2%29 as 2log%2810%2C%282%29%29



2log%2810%2C%283x%2B2%29%29%2F2log%2810%2C%282%29%29-log%2810%2C%28x%29%29%2F2log%2810%2C%282%29%29=3 Multiply the first fraction by 2%2F2



%282log%2810%2C%283x%2B2%29%29-log%2810%2C%28x%29%29%29%2F2log%2810%2C%282%29%29=3 Combine the fractions



%28log%2810%2C%283x%2B2%29%5E2%29-log%2810%2C%28x%29%29%29%2F2log%2810%2C%282%29%29=3 Rewrite 2log%2810%2C%283x%2B2%29%29 as log%2810%2C%283x%2B2%29%5E2%29



%28log%2810%2C%283x%2B2%29%5E2%2Fx%29%29%2F2log%2810%2C%282%29%29=3 Combine the logs



%28log%2810%2C%283x%2B2%29%5E2%2Fx%29%29%2Flog%2810%2C%284%29%29=3 Rewrite 2log%2810%2C%282%29%29 as log%2810%2C%284%29%29



%28log%284%2C%283x%2B2%29%5E2%2Fx%29%29=3 Use the change of base formula again to rewrite the log



4%5E3=%283x%2B2%29%5E2%2Fx Now use the property log%28b%2Cx%29=y ---> b%5Ey=x


64=%283x%2B2%29%5E2%2Fx Raise 4 to the 3rd power to get 64


64x=%283x%2B2%29%5E2 Multiply both sides by x


64x=9x%5E2%2B12x%2B4 Foil


0=9x%5E2-52x%2B4 Subtract 64x from both sides



Let's use the quadratic formula to solve for x:


Starting with the general quadratic

ax%5E2%2Bbx%2Bc=0

the general solution using the quadratic equation is:

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29



So lets solve 9%2Ax%5E2-52%2Ax%2B4=0 ( notice a=9, b=-52, and c=4)




x+=+%28--52+%2B-+sqrt%28+%28-52%29%5E2-4%2A9%2A4+%29%29%2F%282%2A9%29 Plug in a=9, b=-52, and c=4



x+=+%2852+%2B-+sqrt%28+%28-52%29%5E2-4%2A9%2A4+%29%29%2F%282%2A9%29 Negate -52 to get 52



x+=+%2852+%2B-+sqrt%28+2704-4%2A9%2A4+%29%29%2F%282%2A9%29 Square -52 to get 2704 (note: remember when you square -52, you must square the negative as well. This is because %28-52%29%5E2=-52%2A-52=2704.)



x+=+%2852+%2B-+sqrt%28+2704%2B-144+%29%29%2F%282%2A9%29 Multiply -4%2A4%2A9 to get -144



x+=+%2852+%2B-+sqrt%28+2560+%29%29%2F%282%2A9%29 Combine like terms in the radicand (everything under the square root)



x+=+%2852+%2B-+16%2Asqrt%2810%29%29%2F%282%2A9%29 Simplify the square root (note: If you need help with simplifying the square root, check out this solver)



x+=+%2852+%2B-+16%2Asqrt%2810%29%29%2F18 Multiply 2 and 9 to get 18

So now the expression breaks down into two parts

x+=+%2852+%2B+16%2Asqrt%2810%29%29%2F18 or x+=+%2852+-+16%2Asqrt%2810%29%29%2F18


Now break up the fraction


x=%2B52%2F18%2B16%2Asqrt%2810%29%2F18 or x=%2B52%2F18-16%2Asqrt%2810%29%2F18


Simplify


x=26+%2F+9%2B8%2Asqrt%2810%29%2F9 or x=26+%2F+9-8%2Asqrt%2810%29%2F9


So these expressions approximate to

x=5.69980236459412 or x=0.0779754131836626


So our solutions are:
x=5.69980236459412 or x=0.0779754131836626