# Maths logs

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#10

63 = 9*7

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why is it telling me to use base 10?

(i put what i think the answer is on the calculator)

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#12

(Original post by

why is it telling me to use base 10?

(i put what i think the answer is on the calculator)

**A0W0N**)why is it telling me to use base 10?

(i put what i think the answer is on the calculator)

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(Original post by

The right hand side are (mostly) easy to calculate using base 10.

**mqb2766**)The right hand side are (mostly) easy to calculate using base 10.

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#14

(Original post by

but wouldn't they produce different results?

**A0W0N**)but wouldn't they produce different results?

log(0.01) = -2 as 10^(-2) = 0.01

log(0.1) = -1 as 10^(-1) = 0.1

log(1) = 0 as 10^0 = 1

log(10) = 1 as 10^1 = 10

log(100) = 2 as 10^2 = 100

..

Last edited by mqb2766; 6 months ago

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#15

**A0W0N**)

why is it telling me to use base 10?

(i put what i think the answer is on the calculator)

So x=log(10^6)/log(2) which gives the same result. In fact you take logs using any base and the answer above remains the same.

Ofc a mathematician would use log_e

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i understand this question up until the point where I'm told to raise both side to the power of three, i put in log3 (-1) and i got a math error im guessing thats why they're doing something. My question is why has log3 disappeared and -1 has changed to 3^-1 and not -1^3

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#17

(Original post by

i understand this question up until the point where I'm told to raise both side to the power of three, i put in log3 (-1) and i got a math error im guessing thats why they're doing something. My question is why has log3 disappeared and -1 has changed to 3^-1 and not -1^3

**A0W0N**)i understand this question up until the point where I'm told to raise both side to the power of three, i put in log3 (-1) and i got a math error im guessing thats why they're doing something. My question is why has log3 disappeared and -1 has changed to 3^-1 and not -1^3

So raise both sides with a base of 3. The right hand side becomes 3^(-1). The left becomes

3^log(...)

As the log is base 3, this just cancels, leaving the argument which is the term in the brackets.

Last edited by mqb2766; 6 months ago

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(Original post by

Inverse logs is the same as raising the number to the base. The log is base 3 here.

So raise both sides with a base of 3. The right hand side becomes 3^(-1). The left becomes

3^log(...)

As the log is base 3, this just cancels, leaving the argument which is the term in the brackets.

**mqb2766**)Inverse logs is the same as raising the number to the base. The log is base 3 here.

So raise both sides with a base of 3. The right hand side becomes 3^(-1). The left becomes

3^log(...)

As the log is base 3, this just cancels, leaving the argument which is the term in the brackets.

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#19

(Original post by

raising which number to the base?

**A0W0N**)raising which number to the base?

3^log(x^2/(5x+2)) = 3^(-1)

By definition (of a log base 3), the left hand side is just x^2/(5x+2)

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(Original post by

You're raising the expression on each side of the equation. So

3^log(x^2/(5x+2)) = 3^(-1)

By definition (of a log base 3), the left hand side is just x^2/(5x+2)

**mqb2766**)You're raising the expression on each side of the equation. So

3^log(x^2/(5x+2)) = 3^(-1)

By definition (of a log base 3), the left hand side is just x^2/(5x+2)

Is there i video i could watch, and what would i search up if so and how would i enter log3 to the power of three like this:

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