
How do you solve \[20 = 50{\left( {1.04} \right)^x}\] ?
Answer
526.2k+ views
Hint: Here we need to solve for ‘x’. First we isolate the term containing the exponent to one side of the equation. Then we apply logarithm function on both sides of the equation. Afterwards using the logarithmic rules we can solve this. Here we use the power rule of logarithm. That is \[\log {x^a} = a\log x\]
Complete step by step solution:
Given,
\[20 = 50{\left( {1.04} \right)^x}\]
Rearranging we have,
\[50{\left( {1.04} \right)^x} = 20\]
Divide 50 on both side of the equation we have,
\[{\left( {1.04} \right)^x} = \dfrac{{20}}{{50}}\]
\[{\left( {1.04} \right)^x} = \dfrac{2}{5}\]
\[{\left( {1.04} \right)^x} = 0.4\]
Now to get the ‘x’ value we apply the logarithm on both sides of the equation,
\[\log {\left( {1.04} \right)^x} = \log (0.4)\] ,
We have \[\log {x^a} = a\log x\] , applying this we have,
\[x.\log \left( {1.04} \right) = \log (0.4)\]
Divide \[\log \left( {1.04} \right)\] on both sides we have,
\[ \Rightarrow x = \dfrac{{\log (0.4)}}{{\log \left( {1.04} \right)}}\] . This is the required result.
If we want we can continue this simplification using scientific calculator, we will get:
\[ \Rightarrow x = - 23.36241\]
So, the correct answer is “ x = - 23.36241”.
Note: To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is \[\log (x.y) = \log (x) + \log (y)\] . Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. that is \[\log \left( {\dfrac{x}{y}} \right) = \log x - \log y\] . Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is \[\log {x^a} = a\log x\] .
Complete step by step solution:
Given,
\[20 = 50{\left( {1.04} \right)^x}\]
Rearranging we have,
\[50{\left( {1.04} \right)^x} = 20\]
Divide 50 on both side of the equation we have,
\[{\left( {1.04} \right)^x} = \dfrac{{20}}{{50}}\]
\[{\left( {1.04} \right)^x} = \dfrac{2}{5}\]
\[{\left( {1.04} \right)^x} = 0.4\]
Now to get the ‘x’ value we apply the logarithm on both sides of the equation,
\[\log {\left( {1.04} \right)^x} = \log (0.4)\] ,
We have \[\log {x^a} = a\log x\] , applying this we have,
\[x.\log \left( {1.04} \right) = \log (0.4)\]
Divide \[\log \left( {1.04} \right)\] on both sides we have,
\[ \Rightarrow x = \dfrac{{\log (0.4)}}{{\log \left( {1.04} \right)}}\] . This is the required result.
If we want we can continue this simplification using scientific calculator, we will get:
\[ \Rightarrow x = - 23.36241\]
So, the correct answer is “ x = - 23.36241”.
Note: To solve this kind of problem we need to remember the laws of logarithms. Product rule of logarithm that is the logarithm of the product is the sum of the logarithms of the factors. That is \[\log (x.y) = \log (x) + \log (y)\] . Quotient rule of logarithm that is the logarithm of the ratio of two quantities is the logarithm of the numerator minus the logarithm of the denominator. that is \[\log \left( {\dfrac{x}{y}} \right) = \log x - \log y\] . Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is \[\log {x^a} = a\log x\] .
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