Question
Question: How do you solve \({{6}^{x-3}}=52\) ?...
How do you solve 6x−3=52 ?
Solution
We have been asked to solve the expression, 6x−3=52. So, first we will check what type of an equation it is, once we understand that it's an exponential equation, we will try to convert 52 into exponential form. If we are not able to do so, we will try using the log function and then solve using the log function. We will apply log on both the sides, then by using the property of log, that is, log(ab)=bloga, we will solve our question. We will use an algebraic tool to simplify our answer.
Complete answer:
We have been given an equation, 6x−3=52 and are asked to solve it. That means we have to find the value of ‘x’ from the equation, such that both the sides become equal.
Now, first when we consider the equation, we can see that it is in the form of an exponential equation.
So, to solve this, we will first try to convert both the sides into exponential form and compare their powers, and if that is not possible we will use the log function to cancel the exponential function and then solve the equation to get the value of ‘x’.
Now, we have 6x−3 on the left side and 52 on the right side, so we can try to express 52 in the exponential form. We can find the factors of 52. So, we get, 52=2×2×13. From this we understand that it cannot be converted to an exponential form.
Hence, we can use the log function. On applying log on both the sides of the equation, we get,
log6x−3=log52
Now, we know that log(ab)=bloga, so we get,
(x−3)log6=log52
On dividing both the sides by log 6, we get,
(x−3)=log6log52
On simplifying, we get,
x−3=2.205
On adding 3 to both the sides, we get,
x−3+3=2.205+3⇒x=5.205
Hence, we get the value of x as 5.205.
So, our solution is 5.205.
Note: We can compare exponential functions on the right and left side only if their base is the same. Here, since the base of 52 is 2 and 13, and for 6 it's 2 and 3, we cannot compare them. Hence, we can always use the log function to solve the exponential equations. Comparing exponential powers is an easy method, so we always start with that method to solve questions.