Question 636471: Hi, I need help with several questions.
Solve for x.
6^(log_6 of x^2) = x+30 I'm not sure if this makes sense, but that log_6 of x^2 literally shows up as an exponent on my worksheet.
2-log_(2-x)of 2x = 0
I'm not sure if this makes sense either...
Let ln A=8, ln B=-3, and ln C=5. Evaluate each of the following.
1. ln 4 sqrt of A. The 4 is part of the square root.
2. ln (Be)/C
3. ln sqrt of ((A^5Ce^2)/B^3) meaning everything is under the square root.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! Solve for x.
6^(log_6 of x^2) = x+30
Base raised to log of number=number (left side)
x^2=x+30
x^2-x-30=0
(x-6)(x+5)=0
x=-5 (reject, x>0)
x=6
..
2-log_(2-x)of 2x = 0
log_(2-x)of 2x =2
convert to exponential form (base(2-x) raised to log of number(2)=number(2x)
(2-x)^2=2x
4-4x+x^2=2x
x^2-6x+4=0
solve by quadratic formula
x≈5.236.. (reject,(2-x)>0)
or
x≈.764..
..
Let ln A=8, ln B=-3, and ln C=5. Evaluate each of the following.
1. ln 4 sqrt of A. The 4 is part of the square root.
ln√4A
=ln(2*√A)
=ln2+(1/2)lnA
=ln2+(1/2)*8
=ln2+4
=.693+4
=4.693
..
2. ln (Be)/C
=lnB+lne-lnC
=-3+1-5
=-7
..
3. ln sqrt of ((A^5Ce^2)/B^3)
=(1/2)[5lnA+lnC+2lne-3lnB]
=(1/2)[5*8+5+2-3*-3]
=(1/2)[40+7+9]
=[56]/2
=28
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