How to calculate log 5 base 10

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Use our | Log10 calculator to find the logarithm of any positive number for any number base you enter.

What is logarithm?

A logarithm is the power to which a number must be raised in order to get some other number. In other words, the logarithm tells us how many of one number should be multiplied to get another number.

For example:

• The base 2 logarithm of 4 is 2, because 2 raised to the power of 2 is 4:
• log3 9 = 2, because 32 = 9

This is an example of a base-3 logarithm. We call it a base-3 logarithm because 3 is the number that is raised to a power.

The most common logarithms are natural logarithms and base 10 logarithms. There are special notations for them:

• A base 10 log is written simply log.
• A natural logarithm is written simply as ln.

So, the notation log alone means base ten logarithm and notation ln, means natural log.

Basic Log Rules

• logb(x·y) = logb(x) + logb(y)
• logb(x/y) = logb(x) - logb(y)
• logb(xy) = y·logb(x)
• logb(x) = logk(x)/logk(b)

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Precalculus Properties of Logarithmic Functions Common Logs

1 Answer

Rui D.

Apr 16, 2016

See explanation

Explanation:

If you have memorized that
#log2=0.3#
you can follow this way
#log5=log(10/2)=1-log2=1-0.3=0.7#

If you want a general way to find logarithms without using calculators or tables, you could use this formula:
#(1/2)ln|(1+x)/(1-x)|=f(x)=x+x^3/3+x^5/5+...#
And
#logy=lny/ln10=2/ln10*(1/2*ln|y|)#=>#logy=0.869*(1/2*ln|y|)#where#y=(1+x)/(1-x)#
(Note1: you can use#2/ln10= 0.868589#with the precision you like. Using two terms of the series, 0.869 has a proper level of precision. Note 2: the values of x must be smaller than 1.)

We can't calculate#log5#directly because
#(x+1)/(1-x)=5#=>#x+1=5-5x#=>#6x=4#=>#x=1.5#
And the series doesn't converge when#x>1#

But since#5=2*2.5#
for#y_1=2 ->(x+1)/(1-x)=2#=>#x+1=2-2x#=>#x=1/3~=0.3333#
#f(x=1/3)=1/3+1/3^3*1/3=1/3+1/81=0.3333+0.0123=0.3456#

for#y_2=2.5 -> (x+1)/(1-x)=2.5#=>#x+1=2.5-2.5x#=>#3.5x=1.5#=>#x=3/7~=0.4286#
Of course we can use this#x=0.4286#. But perhaps there is an easier way (without a calculator we need to think of this) such as:

Considering that#5=2^2*1.25#(and since we have already calculated#f(x=1/3)#):
for#y_2=1.25 -> (x+1)/(1-x)=1.25#=>#x+1=1.25-1.25x#=>#2.25x=0.25#=>#x=25/225=1/9~=0.1111#
#f(x=1/9)=0.1111+1/9^3*1/3=0.1111+1/729*1/3=1/9+1/2187=0.1111+0.0005=0.1116#
(as to the number#0.0005#just remember that#10/2=5#)

Using the results above
#log5=0.869[2*(1/2*ln|2|)+(1/2*ln|1.25|)]=0.869[2*f(x=1/3)+f(x=1/9)]=0.869[2*0.3456+0.1116]=0.869[0.6912+0.1116]=0.869*0.8028=0.6976332#or#0.698#in 3 decimals

We should be aware that this last estimate is smaller than the correct result.

(In fact#log5=0.6990#)

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