Verified answer

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)

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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

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RE:

Two really difficult log problems?

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...

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

difficult log problems