Question:

Two really difficult log problems?

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I couldn't figure these two out, help please?

If there's a ' v ' before the number, it means that that number's the base

1. Rewrite as a single logarithm

3 log v2 x + 4 log v3 y - (3 log v2 z + 5 log v2 w)

2. Simplify without using base 10 or base e logs, only using the properties of logs. The answer shouldn't be in terms of logs

log v3 [3( fifth root(81) ) ] - log v9 [fourth root(1/ 27)] + 3 log v81 9

For 1, I got log v2 ( xy^4(zw)^15) / 3

And for 2, I got 117 / 40

Pretty sure those two figures are off...

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  1. 1. Rewrite as a single logarithm

    3log[base2](x) + 4log[base3](y) - {3log[base2](z) + 5log[base2](w)}

    = log[base2](x³) + log[base3](y^4) - {log[base2](z³) + log[base2](w^5)}

    = log[base2](x³) + log[base2](y^4)/log[base2](3) - log[base2](z³w^5)

    = log[base2](x³/(z³w^5)) + log[base2](y^4)/log[base2](3)

    __________

    2. Simplify without using base 10 or base e logs, only using the properties of logs. The answer shouldn't be in terms of logs

    log[base3][3(81)^(1/5)] - log[base9][(1/27)^(1/4)] + 3log[base81]9

    = log[base3][3(3^4)^(1/5)] - log[base9][((1/9)^(3/2))^(1/4)] + 3log[base81][(81)^(1/2)]

    = log[base3][3^(9/5)] - log[base9][(1/9)^(3/8)] + 3log[base81][(81)^(1/2)]

    = 9/5 - (-3/8) + 3(1/2) = 9/5 + 3/8 + 3/2 = 147/40


  2. I assume v2 means base 2, v3 means base 3

    (1) rewrite as

    logv2 x^3 + log v3 y^4 - log v2 z^3 - log v2 w^5

    = log v2 (x^3)/(z^3 w^5) + log v3 y^4

    (2) rewrite as

    log v3 3 + (1/5) log v3 81 - log v9 27^(1/4) + log v81 9^3 =

    log v3 3 + (1/5) log v3  81 - (1/4) log v9 3*9 + 3 log v81 9 =

    1 + (1/5)4 - (1/4) log v9 9 - (1/4) log v9 3 + 3 log v81 9 =

    = 1 + (4/5) - (1/4)(1) - (1/4)(1/2) + 3(1/2) =

    I'll leave the arithmetic to you

      

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