2(logbase(3))+6(logbase(x)3)-7=0?

Question

help me to solve this problem in detail.

Answer 1

2 log[base 3](x) + 6 log[base x](3) - 7 =0

Use the log property log[base a](b) = 1 / log[base b](a) (noticed the base and argument are swapped). Let's use this on the second log.

2 log[base 3](x) + 6 (1 / log[base 3](x) ) - 7 = 0

Multiply everything by log[base 3](x).

2 ( log[base 3](x) )^2 + 6 - 7 log[base 3](x) = 0

Reorder this in descending power of log.

2 [log[base 3](x)]^2 - 7 log[base 3](x) + 6 = 0

Which we can factor as a quadratic. It's equivalent to factoring
2y^2 - 7y + 6 = 0 and this factors as (2y - 3)(y - 2), so

( 2 log[base 3](x) - 3 ) ( log[base 3](x) - 2 ) = 0

Giving us two equations.

2 log[base 3](x) - 3 = 0
log[base 3](x) - 2 = 0

Let's solve them both individually:

2 log[base 3](x) - 3 = 0
2 log[base 3](x) = 3
log[base 3](x) = 3/2. Putting this in exponential form,

3^(3/2) = x

log[base 3](x) - 2 = 0
log[base 3](x) = 2
3^2 = x, or x = 9

Therefore, our two solutions are

x = { 3^(3/2) , 9 }

3^(3/2) is equal to (3^3)^(1/2), which is sqrt(27), or 3sqrt(3).

x = { 3sqrt(3), 9 }

Answer 2

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Answer 3

wow that is awesome puggy....I used to be an expert on logs but its been so long I had forgotten some of the rules...thanks for the detailed description