Logarithms... solve for x?

log_4(6 - x) - log_2(8) = log_9(3)
log_4(6 - x) = log_9(3) + log_2(8)
log_4(6 - x) = 0.5 + 3
4^(0.5 + 3) = 6 - x
6 - 4^(3.5) = x
x = 6 - 4^(7/2)
x = 6 - √(4^7)
x = 6 - √(2^7 * 2^7)
x = 6 - 2^7
x = 6 - 128
x = -122

log4 (6 - x) - 3 = 1/2
log4 (6 - x) = 7/2
6 - x = 4^(7/2)
6 - x = 128
x = - 122

this problem cannot be solve at all because the base of the logarithm is not the same.

if only the base are the same like log4(6-x)-log4(8)=log4(3)
this can be solved..

log(4)(6-x)-log(2)(2^3)=log(9)(9^1/2)
log(4)(6-x)-3=1/2
log(4)(6-x)=7/2
6-x=4^(7/2)
x=-122

log(BASE4)(6-x) - log(BASE2)8 = log(BASE9)3.

y=log(BASE9)3 is equivalent to 9^y = 3, so y = .5
z=log(BASE2)8 is equivalent to 2^z = 8, so z = 3

log(BASE4)(6-x) - 3 = .5
log(BASE4)(6-x) = 3.5
4^3.5=6-x
128 = 6-x
x= - 122

i'm not sure if this is correct and i don't have my scientific calculator with me right now so i don't have the final answer.

^ means to the power of

log4 (6 - x) - log2 (8) = log9 (3)
log4 (6 - x) - log2 (2)^3 = log9 (9) ^ 1/2
log4 (6 - x) - 3 = 0.5
log4 (6 - x) = 3.5
6 - x = 4^3.5

i hope you have the answer key cos i haven't done logarithms for quite a while. sorry if it ain't correct!

um ouch my brain hurts i think its um..... nope don-no sorry *crying from massive head ace* T-T

a^x = y is equivalent to log(base a)y=x

log(base2)8=3
log(base9)3=1/2

hence our equation is:
log(base4)6-x - 3 = 0.5
log(base4)6-x = 3.5

thus remembering a^x = y is equivalent to log(base a)y=x

our equation is:
4^3.5=6-x
128=6-x
122 = -x
x = -122

:D

first off

log(2) (8) = 3
log(9) (3) = 1/2

Substitution yields

log(4) (6 - x) + 3 = 1/2
log(4)(6 - x) = -5/2

Note, log(a)(b) = ln(b)/ln(a) --> log(4)(6-x) = ln(6-x)/ln(4)

Subsitutution yields,

ln(6-x)/ln(4) = -5/2
ln(6-x) = -5/2ln(4)
Thus,
6-x = exp(-5/2ln(4))
x = 6 - exp(-5/2ln(4))

Hope this helps,

David

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