SOLUTION: Please help! Thanks a million!!!
Solve the following for X:
1) log_6(2x)=log_6 (2)+log_6 (3x-4)
2) 2ln5=ln(4x-13)
3) log_4(3x+1)=2
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Question 36025: Please help! Thanks a million!!!
Solve the following for X:
1) log_6(2x)=log_6 (2)+log_6 (3x-4)
2) 2ln5=ln(4x-13)
3) log_4(3x+1)=2
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
1) log_6(2x)=log_6 (2)+log_6 (3x-4)
Let's just keep in mind that this is all base 6.
log(2x) = log2 +log(3x-4)
log2x = log2(3x-4)
log2x = log(6x-8)
Take the anti-log of both sides to get:
2x=6x-8
4x=8
x=2
2) 2ln5=ln(4x-13)
ln25=ln(4x-13
25 = 4x-13
4x=38
x=19/2
3) log_4(3x+1)=2
4^2=3x+1
16=3x+1
x=5
Cheers,
stan H.
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