Log Rules
Expand the following:
log_2(xy)^3
3log_2x+3log_2y
Simplify
x^3/x^8
1/x^5
Solve the equation for x
log_2x-log_2 16=0
x=16
Rewrite in exponential form
log_7 (1/49)=-2
7^-2=1/49
How could you rewrite the following?
10logx
logx^10
Expand:
log_4(xy)^2
2log_4x+2log_4y
(a^3b)(ab^6)
a^4b^7
Solve for m
log_3m=3
m=27
Evaluate:
log_x x^(2y)=
=2y
How do I fix this statement to make it true?
For any b>1,
log_9 0=0
By changing the equation to be
log_9 1=0
Expand:
log((2y)/x)
log2+logy-logx
(5a^-3)^2
25/a^6
Solve for k
log_3 (1/9)=k
k=-2
Given
log_4 5=1.2 and log_4 3=0.8, solve log_4 (20/3)
=1.4
What is the solution to and why?
log_4 0=
The solution does not exist because 4^x can't equal 0 for any value of x.
Condense the following using log rules:
2log_2x+4log_2y+5log_2z
log_2x^2y^4z^5
(2fg^4)^4(fg)^6
16f^10g^22
What are the steps to solving this equation?
log_6x=log_6 4+log_6 8
1. Recognize that all the logs have the same base
2. Use product rule to turn the right side into single log
3. Use one to one correspondence, x=32
log_6x=log_6(4*8), log_6x=log_6 32, x=32
Solve for x
4log_3x=2log_2 4
x=2
How do we know that are inverses of each other?
f(x)=10^x and g(x)=logx
The domain of f(x) is the same as the range of g(x) and the range of f(x) is the domain of g(x) OR the x and y values are switched.
Expand using log rules
log_2(2/(xy))^3
3-3log_2x-3log_2y
(2x^3y^-3)^-2
y^6/(4x^6)
Solve
log_2 sqrt32=x
x=5/2
Solve for x
log_3(8x-15)=4
x=12
True or False? (If it's false, fix it to make it true)
log_4 1=4
False,
log_4 4=1