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August 3, 2015

August 3, 2015

Posted by **Nabil** on Saturday, May 19, 2012 at 11:00am.

1.) 2logx-3logy

2.) logx-logy+logz

3.) 1/3log5x+2/3log6x

4.) 2/3(logbase2 x - logbase2 y)

I dont understand these problems at all. Can someone please solve and explain them? Thanks

- Trigonometry -
**Jai**, Saturday, May 19, 2012 at 11:37amTo combine the two log terms, first, the terms must have the same base (the number subscript of log). In this case, if there is no number subscript of log, the base is equal to 10. The two terms have the same base, 10.

Now, recall that to combine log terms of the same base,

*if addition: multiply the terms inside the log. for example

log 3 + log 5 = log (3*5) = log 15

*if subtraction: divide the terms inside the log. for example,

log 3 - log 5 = log (3/5)

*also, if the log is multiplied by a number (outside), we can rewrite it as exponent of the term inside the log. for example,

2log 3 = log (3^2) = log 9

#1.

applying these rules,

2log x - 3log y

we first make the number outside as an exponent to the term inside the log:

log x^2 - log y^3

since subtraction, we divide the terms inside the log:

log (x^2 / y^3)

#2.

log x - log y + log z

log (x/y) + log z

log (xz/y)

#3.

1/3log 5x + 2/3 log 6x

log (5x)^(1/3) + log (6x)^(2/3)

log [(5x)^(1/3) * (6x)^(2/3)]

**recall that to multiply terms with the same base, we add their exponents. for example,

a^(1/3) * a^(2/3) = a^(1/3 + 2/3) = a^(3/3) = a

**also, we can rewrite fraction exponents as, for instance,

2^(2/3) = cuberoot(2^2) = cuberoot (4)

going back to the problem,

log [(5x)^(1/3) * (6x)^(2/3)]

log [5^(1/3) * 6^(2/3) * x^(1/3 + 2/3)]

log [cuberoot(5*6^2) * x^(3/3)]

log [cuberoot(5*36)x]

log [cuberoot(180)x]

#4.

2/3(logbase2 x - logbase2 y)

note that the base now is 2, but that's not a problem since both terms have the same base.

2/3[ logbase2 (x/y) ]

logbase2 (x/y)^(2/3) or

logbase2 [cuberoot (x^2)/(y^2)]

hope this helps~ :)