Solveeit Logo
Login

Question

Question: Given \(\log 4 = .60206\)and \(\log 3 = .4771213\) . Find the logarithms of \(.8,.003,.0108\) and \(...

Given log⁡4=.60206\log 4 = .60206and log⁡3=.4771213\log 3 = .4771213 . Find the logarithms of .8,.003,.0108.8,.003,.0108 and (.00018)17{(.00018)^{\dfrac{1}{7}}} .

Explanation

Solution

Here we are asked to find the logarithm of the given numbers using the given data. To find this, we first need to factorize or modify the given numbers in terms of the given data that is log⁡4\log 4 and log⁡5\log 5 . After that we will substitute the data given that is values of log⁡4\log 4 and log⁡5\log 5 then using some logarithm properties or formulae to find the required result.
Formulas used: Formulae that we need to know:
log⁡a(xn)=nlog⁡ax{\log _a}({x^n}) = n{\log _a}x
log⁡axy=log⁡ax−log⁡ay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y
log⁡a(xy)=log⁡ax+log⁡ay{\log _a}\left( {xy} \right) = {\log _a}x + {\log _a}y

Complete step by step answer:
It is given that log⁡4=.60206\log 4 = .60206 and log⁡3=.4771213\log 3 = .4771213 we aim to find the value of logarithms of .8,.003,.0108.8,.003,.0108 & (.00018)17{(.00018)^{\dfrac{1}{7}}} .
First, let us take the first given value that is 0.80.8 we have to find the logarithm of 0.80.8 that is log⁡0.8\log 0.8
Consider log⁡0.8\log 0.8 , now let’s modify this into the terms of log⁡4\log 4 and log⁡5\log 5.
log⁡0.8=log⁡810\log 0.8 = \log \dfrac{8}{{10}}
=log⁡2310= \log \dfrac{{{2^3}}}{{10}}
Now by using the formula log⁡axy=log⁡ax−log⁡ay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y we get
=log⁡23−log⁡10= \log {2^3} - \log 10
Now let's use the formula log⁡a(xn)=nlog⁡ax{\log _a}({x^n}) = n{\log _a}x on the first term.
=3log⁡2−log⁡10= 3\log 2 - \log 10
This can also be written as
=3log⁡4−log⁡10= 3\log \sqrt 4 - \log 10
=3log⁡412−log⁡10= 3\log {4^{\dfrac{1}{2}}} - \log 10
Now by using the formula log⁡a(xn)=nlog⁡ax{\log _a}({x^n}) = n{\log _a}x again we get
=3×12log⁡4−log⁡10= 3 \times \dfrac{1}{2}\log 4 - \log 10
Now let’s substitute the value of log⁡4\log 4 from the given data and simplify it.
=32(0.60206)−1= \dfrac{3}{2}\left( {0.60206} \right) - 1
=0.90309−1= 0.90309 - 1
⇒log⁡0.8=1‾.90309\Rightarrow \log 0.8 = \overline 1 .90309
Thus, we got the value of the logarithm of the first given number. Let us find the logarithms of other numbers using the same method.
The next given number is 0.0030.003
log⁡0.003=log⁡31000\log 0.003 = \log \dfrac{3}{{1000}}
Now using the formula log⁡axy=log⁡ax−log⁡ay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y we get
=log⁡3−log⁡1000= \log 3 - \log 1000
=log⁡3−log⁡103= \log 3 - \log {10^3}
Now using the formula log⁡a(xn)=nlog⁡ax{\log _a}({x^n}) = n{\log _a}x we get
=log⁡3−3log⁡10= \log 3 - 3\log 10
Now substituting the value of log⁡3\log 3 from the given data and simplifying it we get
=0.4771213−3(1)= 0.4771213 - 3\left( 1 \right)
log⁡0.003=3‾.4771213\log 0.003 = \overline 3 .4771213
Next, we need to find the logarithm of 0.01080.0108
log⁡0.0108=log⁡10810000\log 0.0108 = \log \dfrac{{108}}{{10000}}
Now using the formula log⁡axy=log⁡ax−log⁡ay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y we get
=log⁡108−log⁡10000= \log 108 - \log 10000
=log⁡(4×27)−log⁡104= \log \left( {4 \times 27} \right) - \log {10^4}
=log⁡(4×33)−log⁡104= \log \left( {4 \times {3^3}} \right) - \log {10^4}
Now using the formula log⁡a(xy)=log⁡ax+log⁡ay{\log _a}\left( {xy} \right) = {\log _a}x + {\log _a}y we get
=log⁡4+log⁡33−log⁡104= \log 4 + \log {3^3} - \log {10^4}
Now using the formula log⁡a(xn)=nlog⁡ax{\log _a}({x^n}) = n{\log _a}x we get
=log⁡4+3log⁡3−4log⁡10= \log 4 + 3\log 3 - 4\log 10
Now substituting the value of log⁡4\log 4 & log⁡5\log 5 from the given data we get
=0.60206+3×0.4771213−4×1= 0.60206 + 3 \times 0.4771213 - 4 \times 1
log⁡0.0108=2‾.0334239\log 0.0108 = \overline 2 .0334239
Next, We need to find the logarithm of 0.0001817{0.00018^{\dfrac{1}{7}}} . First, let’s consider the base of the number alone and simplify it.
0.00018=18100000=18105=2×3×3105=2×32105=4×321050.00018 = \dfrac{{18}}{{100000}} = \dfrac{{18}}{{{{10}^5}}} = \dfrac{{2 \times 3 \times 3}}{{{{10}^5}}} = \dfrac{{2 \times {3^2}}}{{{{10}^5}}} = \dfrac{{\sqrt 4 \times {3^2}}}{{{{10}^5}}}
Now let us use this modified term of the number 0.000180.00018
log⁡(0.00018)17=log⁡[4×32105]17\log {\left( {0.00018} \right)^{\dfrac{1}{7}}} = \log {\left[ {\dfrac{{\sqrt 4 \times {3^2}}}{{{{10}^5}}}} \right]^{\dfrac{1}{7}}}
Now using the formula log⁡axy=log⁡ax−log⁡ay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y we get
=17log⁡[4×32105]= \dfrac{1}{7}\log \left[ {\dfrac{{\sqrt 4 \times {3^2}}}{{{{10}^5}}}} \right]
Again, by using the formula log⁡axy=log⁡ax−log⁡ay{\log _a}\dfrac{x}{y} = {\log _a}x - {\log _a}y we get
=17[log⁡(4×32)−log⁡105]= \dfrac{1}{7}\left[ {\log \left( {\sqrt 4 \times {3^2}} \right) - \log {{10}^5}} \right]
Now using the formula log⁡a(xy)=log⁡ax+log⁡ay{\log _a}\left( {xy} \right) = {\log _a}x + {\log _a}y we get
=17[log⁡4+log⁡32−log⁡105]= \dfrac{1}{7}\left[ {\log \sqrt 4 + \log {3^2} - \log {{10}^5}} \right]
Now using the formula log⁡a(xn)=nlog⁡ax{\log _a}({x^n}) = n{\log _a}x we get
=17[12log⁡4+2log⁡3−5log⁡10]= \dfrac{1}{7}\left[ {\dfrac{1}{2}\log 4 + 2\log 3 - 5\log 10} \right]
Now substituting the value from the given data, we get
=17[12×0.60206+2×0.4771213−5×1]= \dfrac{1}{7}\left[ {\dfrac{1}{2} \times 0.60206 + 2 \times 0.4771213 - 5 \times 1} \right]
=17[0.30103+0.9542426−5]= \dfrac{1}{7}\left[ {0.30103 + 0.9542426 - 5} \right]
log⁡(0.00018)17=1‾.4650389\log {\left( {0.00018} \right)^{\dfrac{1}{7}}} = \overline 1 .4650389
Thus, we have found the logarithms of all the given numbers.

Note:
We should note that in decimal numbers as we have in the above solution, the bar over one or more consecutive digits means that the pattern of digits under the bar repeats without end or it is placed in the expression that the expression is to be considered grouped.
It should be noted that if there is no base given in the logarithm expression, then we have to assume that the base is always 1010 . It can be represented as log⁡3=log⁡103\log 3 = {\log _{10}}3 .