need a bit of an explanation on logs

so
5^n-3 = 1/25

i know i have to change the base to 5. thats easy because you dont have to use a calc.
5^n-3 = 5^-2

but...when i get something like 3^2x-1 = 11 i'd have to use the calc and i know the base on calc is 10, but i dont understand the reason why i have to log3 and log11 and do log11/log3? what log law does that follow cos ive covered logs and havent really grasped where that has come from.

but nevertheless without understanding the context, i can just learn the steps and do same type of questions no prob, but im curious as to why i have to do what i have to do to solve it.

Reply 1

B_9710

18

You have

3^{2x-1}=11

so taking logs of both sides ('any' base)

\log(3^{2x-1})=\log 11

and so we have

(2x-1)\log 3=\log 11

using log laws (power law).
Notice if you took logs and used the base as 3 or 11 then either the right hand side or left hand side cancels more nicely. You don't even have to specify the base of the log because it doesn't matter the answer will still be the same.

(edited 7 years ago)

Reply 2

Pikachū

2

Equation isn't clear. Is the 3 in the index or seperate?

Reply 3

TSR George

OP

17

one first thing i thought of which is so obvious: when solving equations, i do the same on both sides, so i log one side then log the other side... am i rite??

but what does log11 or log 3 actually mean? like shouldnt log11 = to something? why is it standing by itself?

Reply 4

NotNotBatman

20

Power rule for logs.

log(x^n) = nlog(x)

Have you learnt that?

Reply 5

B_9710

18

\log_a 11

is the number that satisfies

a^x=11

. That's literally its definition here.

Ok your notation is ambiguous so I am guessing what you intended but here goes ...

3^{2x-1}=11

Take logarithms of both sides. It doesn't matter which base you use but logs to base 10 are usually introduced first. Later on you will almost always use natural logarithms.

\log 3^{2x-1}=\log 11

Now a law of logs is

\log a^b = b \log a

so we use this.

(2x-1) \log 3 = \log 11

Divide both sides by

\log 3

then add 1 and divide by 2.

(edited 7 years ago)

Reply 7

Mistletoe

2

Think of it as the the power you need to raise the smaller number for it to be equal to the larger number.

Reply 8

TSR George

OP

17

Original post by NotNotBatman

Power rule for logs.

log(x^n) = nlog(x)

Have you learnt that?

yes i did, so how did the calculator manage to calculate log11/log3?
is there a nice example that can be done in the head to show how, or its so complicated that only calc can solve it?

Original post by B_9710

You have

3^{2x-1}=11

so taking logs of both sides ('any' base)

\log(3^{2x-1})=\log 11

and so we have

(2x-1)\log 3=\log 11

using log laws (power law).
Notice if you took logs and used the base as 3 or 11 then either the right hand side or left hand side cancels more nicely. You don't even have to specify the base of the log because it doesn't matter the answer will still be the same.

can u show me what it looks like if i use base 3 or 11?

Reply 9

NotNotBatman

20

Well what is

log_{10}11

? It is what we have to raise 10 to get 11. It is the x in

10^x=11

and it's the same with the other log. The example you gave in the op with base 5 is an example for it, but easier as the number is a nice integer.

(edited 7 years ago)

It used something called the CORDIC algorithm or a variant of it. It isn't something you are expected to be able to work out mentally. Even in the olden days before the invention of electronic calculators you'd have needed to look these numbers up in a booklet. You wouldn't know the value of sin (71) without reaching for your calculator either. It's the same difference.

Reply 11

TSR George

OP

17

Original post by NotNotBatman

Well what is

log_{10}11

? It is what we have to raise 10 to get 11. It is the x in

10^x=11

and it's the same with the other log. The example you gave in the op with base 5 is an example for it, but easier as the number is a nice integer.

thanks,

so the example ive given first 5^n-3=1/25 will work if i log it as well?

Reply 12

TSR George

OP

17

Original post by Mr M

It used something called the CORDIC algorithm or a variant of it. It isn't something you are expected to be able to work out mentally. Even in the olden days before the invention of electronic calculators you'd have needed to look these numbers up in a booklet. You wouldn't know the value of sin (71) without reaching for your calculator either. It's the same difference.

ah ok, i guess thats the end of it then, dont have to worry about not knowing more :hoppy:

Reply 13

NotNotBatman

20

What do you mean by it will work? You can find n by taking logs to both sides, because both sides are positive and you "can't" take the log of a negative number.

Yes - try it.

OP

Original post by NotNotBatman

What do you mean by it will work? You can find n by taking logs to both sides, because both sides are positive and you "can't" take the log of a negative number.

dumb worded question, what i meant was i did that question mentally with the knowledge of indices i can change the base. what i couldnt think of was that what i was doing mentally was the same as logs...if you get me, if u dont, dw, i've cracked my confusion now thanks