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Solve this equation: `-2 log_3(x) = log_3(16)` ?

Solve this equation: `-2 log_3(x) = log_3(16)` ?

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Solve this equation: `-2 log_3(x) = log_3(16)` ?

Postby Integr88 » Thu 31 March, 2011 09:47

How to do you solve for x in equation: `-2 log_3(x) = log_3(16)` ?
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Re: Solve this equation: `-2 log_3(x) = log_3(16)` ?

Postby barnamah » Thu 31 March, 2011 11:01

this equations `-2 log_3(x) = log_3(16)` and be written as
we know that `log_a(x)=ln(a)/ln(x)` we apply this rule to simplify the equation


for the left we we have `2 log_3(x)=-2ln(x)/ln(3)`
for the right we have `log_3(16)=ln(16)/ln(3)`. `ln(16)` can be written as `ln(4^2)` and in log or ln we have bring the power down to the left and it becomes `2ln(4)`putting them together we have:


`(-2ln(x))/ln(3)=(2ln(4))/ln(3)`

Cross multiply which mean numerator of one multiply by the denominator of the other and we have:
`-2ln(x)*ln(3)=*2ln(3)*ln(4)*`

Now as you can see we have `2ln(3)` on both side of equation `:.` we can eliminate them (omit them) and we have:
`-ln(x)=ln(4)`

we can now put both side to the power of `e` for example if we `ln(x)=5` we can do `e^(ln(x))=e^5`. Now e and ln one the left side cancel each other and we have `x=e^5`

Applying rule we have:
`e^(-ln(x))=e^(ln(4))`

on the left we have `e^(-ln(x))` which can be written as `1/e^ln(x)` we have:
`e^(-ln(x))=1/e^ln(x)`

Now e cancels the ln on both side and we left with:
`1/x=4`

Now cross-multiply and we have `4x=1` by multiplying both side with `1/4` we have `1/4*4x=1*1/4` so the 4 on the left cancels out and on the right we left with `1/4` which is the answer.


`:.` The final answer is `x=1/4`
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