Super Crazy PreCalculus Problems? Help me!?

Question

5. Simplify: (a) log2 5 + log2(x2 -1) - log2(x - 1)

(b) 2 log4 9 - log2 3

(c) 3^2 log3 5

Answer 1

Basic rules to remember:

LogBaseN(x) + LogBaseN(y) = LogBaseN(xy)

LogBaseN(x) - LogBaseN(y) = LogBaseN(x÷y)

x·LogBaseN(y) = y^x

N^LogBaseN(x) = x

LogBaseN(N^x) = x

LogBaseN(x) = Ln(x) ÷ Ln(N)

Answer 2

{1}-- log_b(a) + log_b(c) = log_b(ac)

{2}-- log_b(d) - logb(f) = log_b(a/f)

a.)------

log_2(5) + log_2(x^2 - 1) - log_2(x - 1)

by the first rule {1}

= log_2(5*[x^2 - 1]) - log_2(x-1)

= log_2(5x^2 - 5) - log_2(x-1)

now by rule {2}

= log_2([5x^2 - 5]/[x-1])

then, if you need to use an old scientific calculator

{3.1}-- log_b(a) = log(a)/log(b)  ÃƒÂ‚    OR if you want to use ln instead

{3.2}-- log_b(a) = ln(a)/ln(b)

b.)------

2log_4(9) - log_2(3)

by rule {3.1} OR {3.2}  --- i'm going to show how to use these rules

= 2[log(9)/log(4)] - log(2)/log(3)

common denominator

= {2[log(9)log(3)] - log(2)log(4)} / {log(4)log(3)}

= {2log(3*3)log(3) - log(2)log(2*2)} / {log(4)log(3)}

by rule {1}

= {2[log(3)+log(3)]log(3) - log(2)[log(2)+log(2)]} / {log(4)log(3)}

= {2[2log(3)]log(3) - log(2)[2log(2)]} / {log(4)log(3)}

= {4[log(3)]^2) - 2[log(2)]^2} / {log(4)log(3)}

= 2 * {[log(3)]^2 - [log(2)]^2} / {log(4)log(3)}

use calculator

2 log_4(9) - log_2(3)

by rule {1}

=2 * log(4)/log(9) - log(2)/log(3)

=1.584962501

c.)----------

3^2 log_3(5)

by rule {1}

= 3^2 log(3)/log(5)

= 4.394920562

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SHARP EL-W531H WriteView  has a function to directly calculate log_b(a). You do not need to convert the logs!

Answer 3

a) assuming 2 is base

and using formula  log X+log Y  = log(X*Y)

    log X - log Y  = log(X/Y)

we get  log ( 5* (x+1)(x-1)) - log(x-1)

             log (5 * (x+1)(x-1)/(x-1))

             log (5 * (x+1))

b)  log (a) b base a

     can be written as   log b / log a

     to any base

so we get

      2 log (4)  9 =   2  log 9/log 4 = 2 log 9/ log 2^2

       = 2 log 9/  2 log 2

      = log 9/ log 2

    ie   log (2)  9

    we get  log  (2)  (9/3)

    log (2)  3

c)

     3^ 2 log 3  5

  =  3 ^ log  3  25

  =   25  (using property a  ^ log a  b  = b)

Answer 4

5.(a) log2 5 + log2(x2 -1) - log2(x - 1)

      =log2(( 5 x (2x-1)) / (x-1))

      =log2((10x-5) / (x-1))

   (b) 2 log4 9 - log2 3

      =2( log2(9) /log2(4) ) - log2(3)

      =2( log2(9)/ log2(2^2) ) - log2(3)

      =2(log2(9) / 2log2(2)) - log2(3)      ***hey, log2(2)=1

      =2log2(9) / 2 - log2(3)

      =log2(9) - log2(3)

      =log2(9/3)

      =log2(3)    **u can stop here**

      =ln3 / ln2

      =1.58496

   (c)3^2 log3 5

      = 3^(log3(5^2))

      = 3^(log3(25))

      = 3^(ln25 / ln3)

      = 3^(2.92995)

      =25