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Question

Question: How do you solve the equation \(\log x = \log 5 + 3\log 3\)?...

How do you solve the equation logx=log5+3log3\log x = \log 5 + 3\log 3?

Explanation

Solution

This problem deals with solving logarithms. Here some basic properties and identities of logarithms are used for simplifying the given equation which is in terms of logarithmic expressions, and linear in the variable xx. The formulas or the properties of logarithms which are used to solve the given problem are:
logab=loga+logb\Rightarrow \log ab = \log a + \log b
nloga=logan\Rightarrow n\log a = \log {a^n}
alogab=b\Rightarrow {a^{{{\log }_a}b}} = b

Complete step-by-step solution:
Consider the given logarithmic equation below as shown:
logx=log5+3log3\Rightarrow \log x = \log 5 + 3\log 3
Now consider the second term of the above equation which is 3log33\log 3 as shown below:
3log3\Rightarrow 3\log 3
Applying the basic property of the logarithms which is nloga=logann\log a = \log {a^n} to the above expression as shown below:
3log3=log33\Rightarrow 3\log 3 = \log {3^3}
Now substituting this simplified above expression in the given logarithmic equation as shown below:
logx=log5+3log3\Rightarrow \log x = \log 5 + 3\log 3
logx=log5+log33\Rightarrow \log x = \log 5 + \log {3^3}
We know that the value of 33=27{3^3} = 27, so substituting this in the above expression as shown below:
logx=log5+log27\Rightarrow \log x = \log 5 + \log 27
Now consider the right hand side of the above equation which is log5+log27\log 5 + \log 27, now applying the basic property of logarithms which is loga+logb=log(ab)\log a + \log b = \log \left( {ab} \right), as shown below:
logx=log5+log27\Rightarrow \log x = \log 5 + \log 27
logx=log5(27)\Rightarrow \log x = \log 5\left( {27} \right)
Now simplifying the above expression as shown below:
logex=loge(135)\Rightarrow {\log _e}x = {\log _e}\left( {135} \right)
So on taking the exponents on both sides of ee, as shown below:
elogex=eloge(135)\Rightarrow {e^{{{\log }_e}x}} = {e^{{{\log }_e}\left( {135} \right)}}
We know the basic property of logarithms which are alogax=x{a^{{{\log }_a}x}} = x, applying this property to the above expression as shown below:
elogex=eloge(135)\Rightarrow {e^{{{\log }_e}x}} = {e^{{{\log }_e}\left( {135} \right)}}
x=135\therefore x = 135
The solution of xx in logx=log5+3log3\log x = \log 5 + 3\log 3 is equal to 135.

Note: Please note that the above problem is solved with the help of applications of some basic identities and properties of logarithms, such as the multiplication rule of logarithms, and the exponential rule of the logarithms. There are few more important properties of logarithms such as division rule and few, such as:
log(ab)=logalogb\Rightarrow \log \left( {\dfrac{a}{b}} \right) = \log a - \log b
The basic logarithmic property is, if logax=b{\log _a}x = b, then the value of x=abx = {a^b}.