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Question

Question: Solve the expression, \[2{{\log }_{10}}5+{{\log }_{10}}8-\dfrac{1}{2}{{\log }_{10}}4.\]...

Solve the expression, 2log105+log10812log104.2{{\log }_{10}}5+{{\log }_{10}}8-\dfrac{1}{2}{{\log }_{10}}4.

Explanation

Solution

Simplify the above expression using the basic rules of logarithm. Use the log of power rule, product rule, and quotient rule to simplify and find the value of the expression.

Complete step-by-step answer:
In mathematics, the logarithm is the inverse function to exponentiation. That means the logarithm of a given number x is the exponent to which another fixed number, the base b must be raised to produce that number x.
The common logarithm is the logarithm to base 10. The common logarithm of x is the power to which the number 10 must be raised to obtain the value of x. Thus the common logarithm of 10 is 1.
We have been given 2log105+log10812log104.2{{\log }_{10}}5+{{\log }_{10}}8-\dfrac{1}{2}{{\log }_{10}}4.
Now let us apply the basic logarithm’s formula and solve the expression.
logab=bloga{{\operatorname{loga}}^{b}}=b\log a, by the log of power.
Thus we can write 2log105=log10522{{\log }_{10}}5={{\log }_{10}}{{5}^{2}} and 1/2log104=log1041/2=log104=log102{}^{1}/{}_{2}{{\log }_{10}}4={{\log }_{10}}{{4}^{{}^{1}/{}_{2}}}={{\log }_{10}}\sqrt{4}={{\log }_{10}}2.
Thus 2{{\log }_{10}}5+{{\log }_{10}}8-\dfrac{1}{2}{{\log }_{10}}4={{\log }_{10}}{{5}^{2}}+{{\log }_{10}}8-{{\log }_{10}}2$$$$={{\log }_{10}}25+{{\log }_{10}}8-{{\log }_{10}}2.
By the product rule of logarithm log (xy) = log x + log y.
Thus applying this rule to (log1025+log108)\left( {{\log }_{10}}25+{{\log }_{10}}8 \right) we get,
log1025+log108log102=log10(25×8)log102=log10(200)log102.{{\log }_{10}}25+{{\log }_{10}}8-{{\log }_{10}}2={{\log }_{10}}\left( 25\times 8 \right)-{{\log }_{10}}2={{\log }_{10}}\left( 200 \right)-{{\log }_{10}}2.
By quotient rule of algorithm, logxlogy=log(x/y)\log x-\operatorname{logy}=log\left( {}^{x}/{}_{y} \right).
Thus applying this rule in the above expression, we get,
log10200log102=log10(2002)=log10100\Rightarrow {{\log }_{10}}200-{{\log }_{10}}2={{\log }_{10}}\left( \dfrac{200}{2} \right)={{\log }_{10}}100.
We know 100=102100={{10}^{2}}
log10100=log10102\therefore {{\log }_{10}}100={{\log }_{10}}{{10}^{2}}
Thus by using rule of log power, log(xy)=ylogx\log \left( {{x}^{y}} \right)=y\operatorname{logx}
log10102=2log1010lo{{g}_{10}}{{10}^{2}}=2{{\log }_{10}}10
We know that log1010=1{{\log }_{10}}10=1
2log1010=2\therefore 2{{\log }_{10}}10=2.
Thus we got the value of 2log105+log10812log1042{{\log }_{10}}5+{{\log }_{10}}8-\dfrac{1}{2}{{\log }_{10}}4 as 2.

Note: You should know the basic rules of logarithms to solve the above expression. Logarithm is an important concept in mathematics and you should be able to solve problems by using the basic concepts of logarithm.