Solving logs by formula and not a calculator.

log _b(x) = log _c(x) / log _c(b)

so

log ₁₀(x) = log _c(x) / log _c(10)

That isn't that clear to me Rom. Also I'm unsure which part of the question you're answering specifically. Though I feel it's conversion process for any log function into a base 10 log. But it seems alien and a bit backwards, though I'm sure it's not.
Specifically, this does not expalin to me how it goes from a log 4 function to a log 10 function, or how to calculate a log 4 function manually, in steps (since I have the problems and the fomulae but not the method) and It does not really explain what is x b and c, where we have one equation, what's what.

logb(x) = logc(x) / logc(b)

so

log10(x) = logc(x) / logc(10)
If you could explain it a bit more thanks.

My interpretation would be logb is log input any base and it will be calculated by log original base by log new base. And what about powers? And how do we solve for powers? What's the step from log 2 or 4 or 7 etc.


Still would like to know the calculation method to manually calculate for example log ₄78. and maybe you can show a little bit of a more difficult log problem and solve to show me also. But I'd like to know the method of working through it specifically and where powers are involved. Ie. what's being Sqrt'ed, multiplied, divided etc to come up with an answer if any one would like to help.

as your concrete example

log ₄(78) = log ₁₀(78) / log ₁₀(4)

Now you talk about calculating these things manually. I'm not entirely sure what you mean by that. Even for base 10 logs eventually you're going to be reduced to one of 3 ways of getting the actual value

a) punching the buttons on your calculator
b) looking the value up in a table
c) approximating it by it's series or perhaps some other approximation formula

unless of course the argument is a power of the base in which case the log is just that power, for example

log ₁₀(1000) = log ₁₀(10 ³) = 3
log ₄(1024) = log ₄(4 ⁵) = 5

without using one of the methods a-c above about the best you can do is note that

4 ³ = 64 < 78 < 256 = 4 ⁴

and thus 3 < log ₄(78) < 4

back in the old days before electronic calculators (and yes I was alive back then) all we had was tables of base 10 logs and every problem involving logs first had to be converted to base 10 using the formula I gave you, and then done in base 10 and converted back if necessary. With pencil and paper (or a slide rule if you had one).

If you're interested in the series I'm referring to in (c) above you can read about it here. http://en.wikipedia.org/wiki/Logarithm#Calculation

Thanks. (STU)

solve log ₄(1/4).. What do I do with the fraction in a log situation. How in the instance described do I apply log _am ⁿ = n log _am. Oh It should be so easy. Got to work out some of the nuts and bolts I think with this one.

and working this to: 2 log _a5 + log _ap

Quote:

log4(1/4)

*cough*

1/4 = 4 ^(-1)

so log ₄(1/4) = -1

Stu:

solve log₄(1/4).. What do I do with the fraction in a log situation. How in the instance described do I apply log_amⁿ = n log_am. Oh It should be so easy. Got to work out some of the nuts and bolts I think with this one.

and working this to: 2 log_a5 + log_ap

Hi Stu,
Remember this stuff that we did before Stu.
http://web2.0calc.com/questions/viewtopic.php?f=2&t=8007&p=18073&hilit=negative+indices#p18073
take a look and revise.

now 1/4 = 1/4 ¹ ==> 4 ^-1
Remember that Stu? Probably not, I know that you have been swamped.
so
log ₄(1/4) = log ₄(4 ^-1) ==> -1*log ₄4 ==> -1*1 ==> -1
------------------------------------------------------------------------------------------------

2 log _a5 + log _ap =

= log _a5 ² + log _ap

= log _a25 + log _ap

= log _a(25p)

I always keep this page handy when i am doing logs
http://en.wikipedia.org/wiki/List_of_logarithmic_identities

Yeh, I knew it was simple. lol. I even went over this yesterday in a different example to flip the fraction which came to me after posting. Thanks for the reply. lol. Still have to absorb it, enough to remember all, apply as needed and enough to do method in reverse. It's a slow process when it's all new.

Nice wiki link.

Stu:

Yeh, I knew it was simple. lol. I even went over this yesterday in a different example to flip the fraction which came to me after posting. Thanks for the reply. lol. Still have to absorb it, enough to remember all, apply as needed and enough to do method in reverse. It's a slow process when it's all new.

Nice wiki link.

We know it is hard for you Stu. There is so much that you have to learn.