What is log5 (25x25/5)=

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Posted on Jan 02, 2017

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To calculate the log base b of x, divide the log of x by the log of b. You can use either the common log (log) or natural log (ln) as long as you use the same one for both.

For example, to calculate the log base 5 of 10, divide

log 10 by log 5

or

ln 10 by ln 5 .

Either way you should get about 1.43 .

For example, to calculate the log base 5 of 10, divide

log 10 by log 5

or

ln 10 by ln 5 .

Either way you should get about 1.43 .

Apr 12, 2013 | Casio Prizm FX-CG10 Color Graphing...

Type **solve(log(x^5,16)-log(125,64)+log(x^(1/2),4),x,0.5**

Its equation with guess 0,5. Then press enter and you will give result

x=5^(1/3)=1.70997

Or, you can use equation solver with eqn=0(MATH, then B option)

See captured images below

x=5^(1/3)=1.70997

Or, you can use equation solver with eqn=0(MATH, then B option)

See captured images below

Feb 26, 2013 | Texas Instruments TI-83 Plus Calculator

+4917677424185 Email Transcript [HVD] ====== Start time: 1/5/2012, 2:12:7 [HVD] Module HVD Log Version = 2,12,32,426 [HVD] OS Info 6 0 S 2 0 [VC] ====== Start time: 1/5/2012, 2:12:7 [VC] Module VC Log Version = 1,7,15,426 [VC] OS Info 6 0 S 2 0 [VC] ====== Start time: 1/5/2012, 2:12:7 [VC] Module VC Log Version = 1,7,15,426 [VC] OS Info 6 0 S 2 0 [YP] ====== Start time: 1/5/2012, 2:12:7 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [YP] GetYoutubeURLType() = -1, http://www.facebook.com/people/Patricia-Geiss/100002655619425 [YP] ====== Start time: 1/5/2012, 2:12:7 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [HVD] ====== Start time: 1/5/2012, 2:12:9 [HVD] Module HVD Log Version = 2,12,32,426 [HVD] OS Info 6 0 S 2 0 [YP] GetYoutubeURLType() = -1, info@web.de [YP] GetYoutubeURLType() = -1, info@web.de [YP] ====== Start time: 1/5/2012, 2:12:12 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [YP] InternetCrackUrl() = 12006, info@web.de [YP] WinHttpCrackUrl() = 12006, info@web.de [YP] GetYoutubeURLType() = -1, info@web.de [YP] GetYoutubeURLType() = -1, info@web.de [YP] ====== Start time: 1/5/2012, 2:12:51 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [YP] InternetCrackUrl() = 12006, info@web.de [YP] WinHttpCrackUrl() = 12006, info@web.de [HVD] ====== Start time: 1/5/2012, 2:14:48 [HVD] Module HVD Log Version = 2,12,32,426 [HVD] OS Info 6 0 S 2 0 [VC] ====== Start time: 1/5/2012, 2:14:48 [VC] Module VC Log Version = 1,7,15,426 [VC] OS Info 6 0 S 2 0 [VC] ====== Start time: 1/5/2012, 2:14:48 [VC] Module VC Log Version = 1,7,15,426 [VC] OS Info 6 0 S 2 0 [YP] ====== Start time: 1/5/2012, 2:14:48 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [YP] GetYoutubeURLType() = -1, http://www.facebook.com/people/Patricia-Geiss/100002655619425 [YP] ====== Start time: 1/5/2012, 2:14:48 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [HVD] ====== Start time: 1/5/2012, 2:14:49 [HVD] Module HVD Log Version = 2,12,32,426 [HVD] OS Info 6 0 S 2 0 [YP] GetYoutubeURLType() = -1, snc4/203372_100002655619425_1800818_n.jpg [YP] GetYoutubeURLType() = -1, snc4/203372_100002655619425_1800818_n.jpg [YP] ====== Start time: 1/5/2012, 2:14:51 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [YP] InternetCrackUrl() = 12006, snc4/203372_100002655619425_1800818_n.jpg [YP] WinHttpCrackUrl() = 12006, snc4/203372_100002655619425_1800818_n.jpg [YP] GetYoutubeURLType() = -1, snc4/203372_100002655619425_1800818_n.jpg [YP] GetYoutubeURLType() = -1, snc4/203372_100002655619425_1800818_n.jpg [YP] ====== Start time: 1/5/2012, 2:15:11 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [YP] InternetCrackUrl() = 12006, snc4/203372_100002655619425_1800818_n.jpg [YP] WinHttpCrackUrl() = 12006, snc4/203372_100002655619425_1800818_n.jpg [YP] GetYoutubeURLType() = -1, snc4/203372_100002655619425_1800818_n.jpg [YP] GetYoutubeURLType() = -1, snc4/203372_100002655619425_1800818_n.jpg [YP] ====== Start time: 1/5/2012, 2:15:15 [YP] Module YP Log Version = 2,12,32,426 [YP] OS Info 6 0 S 2 0 [YP] InternetCrackUrl() = 12006, snc4/203372_100002655619425_1800818_n.jpg [YP] WinHttpCrackUrl() = 12006, snc4/203372_100002655619425_1800818_n.jpg

Feb 09, 2012 | Health & Beauty

You calculate the pH with the relation

pH=-log(c) where c is the value of concentration in mol/L.

ex: c[H+]= 3.57x10^(-6)

pH: (-) [LOG] 3.57 [2nd][EE] [(-)] 6 [ ) ]

Here you have 2 change sign key press [(-)]

If you have the pH, and want to calculate the concentration that yields that value of the pH you proceed as follows

c= 10^(-pH)

Ex pH=8.23

c: [2nd][LOG] [(-) 8.23 [)]

General shortcut.

If concentration is of the form c[H+]= 1*10^(-a) where a is an integer between 0 and 14, the pH is equal to the positive value of exponent.

Ex:c[H+]=1x10^(-5) Mol/L then the pH is just the positive value of the exponent, ie 5 because pH=-log(1.x10^-5) =- log1 -log(10^-5)=0 -(-5)=5

Above I use the rule log(axb)=log(a)+log(b) , the rule log(a^b) = b*log(a)and the facts that log(1)=0 and log(10)=1, the last being true if log stands for log in base 10.

pH=-log(c) where c is the value of concentration in mol/L.

ex: c[H+]= 3.57x10^(-6)

pH: (-) [LOG] 3.57 [2nd][EE] [(-)] 6 [ ) ]

Here you have 2 change sign key press [(-)]

If you have the pH, and want to calculate the concentration that yields that value of the pH you proceed as follows

c= 10^(-pH)

Ex pH=8.23

c: [2nd][LOG] [(-) 8.23 [)]

General shortcut.

If concentration is of the form c[H+]= 1*10^(-a) where a is an integer between 0 and 14, the pH is equal to the positive value of exponent.

Ex:c[H+]=1x10^(-5) Mol/L then the pH is just the positive value of the exponent, ie 5 because pH=-log(1.x10^-5) =- log1 -log(10^-5)=0 -(-5)=5

Above I use the rule log(axb)=log(a)+log(b) , the rule log(a^b) = b*log(a)and the facts that log(1)=0 and log(10)=1, the last being true if log stands for log in base 10.

May 17, 2011 | Texas Instruments TI-30XA Calculator

You can calculate the log of x to any base b by calculating log(x)/log(b) or ln(x)/ln(b). So, to calculate log 5 to the base 3, press

LOG 5 ) / LOG 3 ENTER

or

LN 5 ) / LN 3 ENTER

LOG 5 ) / LOG 3 ENTER

or

LN 5 ) / LN 3 ENTER

Jan 25, 2011 | Texas Instruments TI-83 Plus Calculator

You calculate the pH with the relation

pH=-log(c) where c is the value of concentration in mol/L.

ex: c[H+]= 3.57x10^(-6)

pH: (-) [LOG] 3.57 [2nd][EE] [(-)] 6 [ ) ]

Here you have 2 change sign key press [(-)]

If you have the pH, and want to calculate the concentration that yields that value of the pH you proceed as follows

c= 10^(-pH)

Ex pH=8.23

c: [2nd][LOG] [(-) 8.23 [)]

General shortcut.

If concentration is of the form c[H+]= 1*10^(-a) where a is an integer between 0 and 14, the pH is equal to the positive value of exponent.

Ex:c[H+]=1x10^(-5) Mol/L then the pH is just the positive value of the exponent, ie 5 because pH=-log(1.x10^-5) =- log1 -log(10^-5)=0 -(-5)=5

Above I use the rule log(axb)=log(a)+log(b) , the rule log(a^b) = b*log(a)and the facts that log(1)=0 and log(10)=1, the last being true if log stands for log in base 10.

pH=-log(c) where c is the value of concentration in mol/L.

ex: c[H+]= 3.57x10^(-6)

pH: (-) [LOG] 3.57 [2nd][EE] [(-)] 6 [ ) ]

Here you have 2 change sign key press [(-)]

If you have the pH, and want to calculate the concentration that yields that value of the pH you proceed as follows

c= 10^(-pH)

Ex pH=8.23

c: [2nd][LOG] [(-) 8.23 [)]

General shortcut.

If concentration is of the form c[H+]= 1*10^(-a) where a is an integer between 0 and 14, the pH is equal to the positive value of exponent.

Ex:c[H+]=1x10^(-5) Mol/L then the pH is just the positive value of the exponent, ie 5 because pH=-log(1.x10^-5) =- log1 -log(10^-5)=0 -(-5)=5

Above I use the rule log(axb)=log(a)+log(b) , the rule log(a^b) = b*log(a)and the facts that log(1)=0 and log(10)=1, the last being true if log stands for log in base 10.

May 20, 2010 | Texas Instruments TI-30XA Calculator

( 3 * 5 ) LOG for common log or ( 3 * 5 ) LN for natural log, assuming you mean log of (3 times 5). If you want (log of 3) times 5 then its 3 LOG x 5 = or 3 LN x 5 =

May 03, 2010 | Texas Instruments TI-30XA Calculator

Use the relation log5(x) = log(x) / log(5) = ln(x) / ln(5)

LOG ( 1 / 625 ) / LOG 5 =

or

LN ( 1 / 625 ) / LN 5 =

LOG ( 1 / 625 ) / LOG 5 =

or

LN ( 1 / 625 ) / LN 5 =

Mar 11, 2010 | Texas Instruments TI-30XA Calculator

You cannot do it with a single key. Rather you use the relation

log in base a of b = log_a(b)=log(b) /log(a)

So log2 5 =log(5)/log(2)

log in base a of b = log_a(b)=log(b) /log(a)

So log2 5 =log(5)/log(2)

Feb 26, 2010 | Casio FX-260 Calculator

to do logarithms with other bases, use change-of-base.
let's say you want to do log base 5 of 25. for clarity i'll write this as log(5) 25.
you would have to do log(10) 25 / log(10) 5 to get 2.
key sequence:
[LOG], 2, 5, ), / (divide), [LOG], 5, )

Apr 03, 2009 | Texas Instruments TI-83 Plus Calculator

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