Question

How do you solve ${3^b} = 17$ ?

Answer 1

Hint : We need to solve for ‘b’. We apply logarithm on both sides of the equation and we apply the power rule of $\log {x^a} = a\log x$ . After solving the equation for ‘b’ we use a calculator or logarithmic table to obtain the value of ‘b’.

Complete step by step solution:
We have,
${3^b} = 17$
Apply logarithm function on both sides we have,
$\log ({3^b}) = \log (17)$ .
Power rule of logarithm that is the logarithm of an exponential number is the exponent times the logarithm of the base. That is $\log {x^a} = a\log x$ .
Where $a = b,x = 3$ .
$b\log (3) = \log (17)$
Divide by $\log (3)$ on both sides we have,
$\Rightarrow b = \dfrac{{\log (17)}}{{\log (3)}}$ .
We know that $\log \left( {17} \right) = 1.2304$ and $\log \left( 3 \right) = 0.4771$
$\Rightarrow b = \dfrac{{1.2304}}{{0.4711}}$
$\Rightarrow b = 2.611$ . This is the required answer.
So, the correct answer is “b = 2.611”.

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 $\log {x^a} = a\log x$ .