Question

How do you solve for x in ${{\log }_{x}}32=5$

Answer 1

Hint: Now we know that the equation ${{\log }_{a}}b=x$ can be written as ${{a}^{x}}=b$ using the definition of the function. Hence we will write the equation in this form and then factorize 32 such that we get 5 equal factors. Hence we get the solution to the equation.

Complete step by step solution:
Now first let us understand the concept of logarithm.
In simple Logarithm is nothing but the power to which a number should be raised to obtain a certain number.
Now let us consider a log term say ${{\log }_{a}}b$ here is called the base of logarithm.
Now the value of ${{\log }_{a}}b$ is nothing but what power must be raised so that we get b.
Hence if we have ${{\log }_{a}}b=x$ then we can say that ${{a}^{x}}=b$ .
Hence now we know the definition of log.
Note that sometimes the base of log is not written, in such case it is assumed to be 10. Also logarithm with base as e is called natural logarithm and is denoted by ln.
Now consider the given equation.
We have ${{\log }_{x}}32=5$ hence by definition we can say that ${{x}^{5}}=32$ .
Now let us factorize 32. Hence we get 32 = 2 × 2 × 2 × 2 × 2
Which means we can say that ${{2}^{5}}=32$
Hence on comparing the two equations we get the value of x = 2.

Note:
Now note that in logarithm the base of the function as well as the variable is always positive and the base of the function is never 1. Also note that the value of log is positive for variables greater than 1 and negative for variables less than 1. Also ${{\log }_{a}}1=0$ .