College Algebra help? Logarithm help?

a) If the bases are the same you can add and subtract the numbers as if they were just that. So for the first one, log3 5 + log3 16 - log3 8 would be the same as log3 (5+16-8) or log3 13.

b.) 4ln x - 1-3ln (y + 2) = ln x^4 - ln (y + 2)^(1/3) because you take the number in front of the natural log and transfer it to the exponent of what the natural log is being multiplied by. Then you combine the natural logs into one long (remember that subtraction means divide and addition means multiply). Your resulting natural log would be ln (x^4)/ [(y + 2)^(1/3)]. From this step it is solvable, but it says you only need the single logarithm. Hope this helps. :)

Write each of the following as a single logarithm.

a) log3 5 + log3 16 - log3 8= log(3){5*16/8} =log(3){10}......Ans

b) 4ln x - 1/3ln (y + 2)= ln(x)^4 - ln(y+2)^(1/3)= ln{x^4/(y+2)^1/3}.....Ans

a)

log3 5 + log3 16 - log3 8 = LOG3[5+16-8] = log3(13) >======< ANSWER

b)

4ln x - 1/3ln (y + 2) = ln(x^4) - ln[(y+2)^(1/3)]

4ln x - 1/3ln (y + 2) = ln[(y +2)^91/3)] >================< ANSWER

a. log3 (5*16/8)

or log3 10

b. 4ln x - 1/3ln ( y+2)

ln x^4 - ln (y+2)^1/3

ln ((x^4)/(y+2)^1/3

a)

Let log be log to base 3

x = log 5 + log 16 - log 8

x = log 5 + log (16/8)

x = log 5 + log 2

x = log 10

b)

Will assume that question should be shown as :-

4 ln x - (1/3) ln ( y + 2 )

ln x^4 + ln ( y + 2 )^(-1/3)

ln [ x^4 / ( y + 2 )^(1/3) ]

log(5*16/8) = log(10), where this log is to base 3.

LN( x^4 / (y+2)^3 )

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