Can you please explain your problem further so we can help you better, because your concern is not that clear....

Posted on Feb 14, 2009

Use another memory stick instead of Mexell.

Posted on Feb 14, 2009

Please explain some more.

thanks

Posted on Feb 13, 2009

Hi,

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Posted on Jan 02, 2017

1 EE 14 LOG / 2 +<>- =

Jun 25, 2010 | Texas Instruments TI-30XA Calculator

I calculators __you can't__ change or set the base of 'log' is __default 10__ so you need to know these:

log x (base y) = log x / log y

when log x it's automatic log x (base 10) here some examples:

log 8 (base 2) = log 8 / log 2

log 27 (base 3) = log 27 / log 3

and the base it's default 10 again, and that is how you need to use it in calculators(no bases), and here the explanation to this:

log x (base y) = log x (base 10) / log y (base 10) = (log x / log 10) / (log y / log 10) = (log x / log 10) * (log 10 / log y) = log x / log y

here so other stuff about 'log' :

log (x*y) = log x + log y

log (x/y) = log x - log y

Hope it's helped**☺**

log x (base y) = log x / log y

when log x it's automatic log x (base 10) here some examples:

log 8 (base 2) = log 8 / log 2

log 27 (base 3) = log 27 / log 3

and the base it's default 10 again, and that is how you need to use it in calculators(no bases), and here the explanation to this:

log x (base y) = log x (base 10) / log y (base 10) = (log x / log 10) / (log y / log 10) = (log x / log 10) * (log 10 / log y) = log x / log y

here so other stuff about 'log' :

log (x*y) = log x + log y

log (x/y) = log x - log y

Hope it's helped

on Jul 07, 2009 | Casio fx-300ES Calculator

Hi,

It is easier to use the rules for logarithms to clean this equation before trying to solve it. And the calculator is no help here.

Let me represent the logs as follows log b in base a = log(a,b). The first argument is the base.

2*log(4,x) - log(4,3)= log(4,x)- log(4,2)

Here you have two similar terms 2*log(4,x) on the left and log(4,x) on the right. Gather them both on the same side, say the left. You have

**2*log(4,x) -log(4,x) +log(4,3) = log(4,x)-log(4,x) -log(4,2) **

which is equivalent to**log(4,x) +log(4,3) = -log(4,2)**

Gather the constant terms on the right, and use log(a) +log(b)=log(a*b)

log(4,x) = -log(4,2) -log(4,3) =**- { log(4,2)+log(4,3)}** = -log(4,2*3)

You get**log(4,x)= -log(4,6)**.

Take the constant term to the left, while changing its sign in the process. On the right remains 0, which is also the log of 1 in any base So log(4,x)+log(4,6) =0 =log(4,1)

Use the rule log(a) +log(b) = log(a*b) valid in any base, to obtain

log(4,6x)=log(4,1) which is equivalent to 6x=1. The solution is x=1/6

There are a few other shortcuts to get the solution but I preferred to take a pedestrian, step by step, approach.

Hope it helps.

It is easier to use the rules for logarithms to clean this equation before trying to solve it. And the calculator is no help here.

Let me represent the logs as follows log b in base a = log(a,b). The first argument is the base.

2*log(4,x) - log(4,3)= log(4,x)- log(4,2)

Here you have two similar terms 2*log(4,x) on the left and log(4,x) on the right. Gather them both on the same side, say the left. You have

which is equivalent to

Gather the constant terms on the right, and use log(a) +log(b)=log(a*b)

log(4,x) = -log(4,2) -log(4,3) =

You get

Take the constant term to the left, while changing its sign in the process. On the right remains 0, which is also the log of 1 in any base So log(4,x)+log(4,6) =0 =log(4,1)

Use the rule log(a) +log(b) = log(a*b) valid in any base, to obtain

log(4,6x)=log(4,1) which is equivalent to 6x=1. The solution is x=1/6

There are a few other shortcuts to get the solution but I preferred to take a pedestrian, step by step, approach.

Hope it helps.

Nov 25, 2009 | Casio FX-300MS Calculator

It's simple here some logs:

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

you just need to know that the default base of log in calculators is 10, so log x mean

log x base 10 = log x / log 10,

so write log x / log y for log x base y example:

log 8 base 2 = log 8 / log 2 that is how you write it on calculator, hope it was helpful

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

you just need to know that the default base of log in calculators is 10, so log x mean

log x base 10 = log x / log 10,

so write log x / log y for log x base y example:

log 8 base 2 = log 8 / log 2 that is how you write it on calculator, hope it was helpful

Jul 01, 2009 | Texas Instruments TI-83 Plus Calculator

You can't it's default 10 so learn this logs:

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

so just divide the logs

hope it helps

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

so just divide the logs

hope it helps

Jun 28, 2009 | Texas Instruments TI-83 Plus Calculator

It's very simple: the calculators have a default base 10 so that log x it's log x base 10 so here is some logs:

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

Hope it's helped

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

Hope it's helped

Apr 09, 2009 | Sharp EL-520WBBK Calculator

Look the calculators has a default base 10 and you can't change it but you may learn this:

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

so instead input log 8 base 2 you need to input log 8 / log 2

your example answer is about 0.919... if i get right what did you write.

Glad to help :)

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

so instead input log 8 base 2 you need to input log 8 / log 2

your example answer is about 0.919... if i get right what did you write.

Glad to help :)

Feb 21, 2009 | HP 33s Calculator

I don't get it but here some help with logs:

log x base y = log x / log y (when log x is default base 10 in calculators) and

log (x*y) = log x + log y

log (x/y) = log x - log y

the bases is default 10

hope it's helped

log x base y = log x / log y (when log x is default base 10 in calculators) and

log (x*y) = log x + log y

log (x/y) = log x - log y

the bases is default 10

hope it's helped

Feb 05, 2009 | Casio FX-300MS Calculator

All calculators it's the same:

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

the default base in all the calculators is 10 so instead of writing log 4 base 2 you simply need to write log 4 / log 2

Hope you understand :)

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

the default base in all the calculators is 10 so instead of writing log 4 base 2 you simply need to write log 4 / log 2

Hope you understand :)

Dec 16, 2008 | Casio FX-7400G Plus Calculator

Here something that would help you:

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

so log 24 base 2 it's log 24 / log 2 etc.

and if i get right what you wrote x= 10 ^ ( 3 * (log 5) ) if no inverse it it's easy really Hope it's helped you.

log x base y = log x / log y

log (x*y) = log x + log y

log (x/y) = log x - log y

so log 24 base 2 it's log 24 / log 2 etc.

and if i get right what you wrote x= 10 ^ ( 3 * (log 5) ) if no inverse it it's easy really Hope it's helped you.

Oct 23, 2008 | Casio FX-260 Calculator

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