Tag Archives: precalculus

Basic Algebra for IITJEE Main and RMO

More basic algebra for you guys who are thirsting for more…The following is a nice problem indicating some basic concepts or tricks in problems involving logarithms/powers.

Solve for x: 4^{x}-3^{x-(1/2)}=3^{x+(1/2)}-2^{2x-1} (IITJEE 1978)

Solution:

Writing 4^{x} as 2^{2x} and bringing powers of the same number on the same side we get,

2^{2x}+2^{2x-1}=3^{x+(1/2)}+3^{x-(1/2)}

The first term on the LHS can be written as 2^{2x-1} \times 2, and hence, a common factor of

2^{2x-1} comes out from the terms on the LHS. As for RHS, we can rewrite the first term as

3^{x-(1/2)} \times 3 and then the factor 3^{x-(1/2)} comes out as common. So, we get

2^{2x-1}(2+1)=3^{x-(1/2)}(3+1), that is, 3 \times 2^{2x-1}=4 \times 3^{x-(1/2)}

Bringing all powers of 2 to  the left and all powers of 3 to the right, we get

2^{2x-3}=3^{x-(3/2)}

By inspection, x=3/2 is a solution. But, how do we arrive at it systematically? Also, how do we know that there is no other solution? It is tempting to try to do this by saying that a power of 2 can equal a power of 3 only when are both are equal to 1. (Such a reasoning is indeed useful in solving equations in Number Theory where we mostly deal with positive integers and their factorization into integers). But, here it is inapplicable because we do not  know that the exponents are integers. Instead, let us express both the sides as the power of the same number. One way to do this  is to write 3 as 2^{\log_{2}3} in the RHS. Then, we can get

2^{2x-3}=2^{(\log_{2}3)(x-(3/2)}

As the bases are the same, the equality of powers implies that of the exponents. So, we have

2x-3=(\log_{2}3)(x-(3/2))

This can be solved easily to give x=\frac{3-(3/2)(\log_{2}3)}{2-\log_{2}3}

which simply equals 3/2.

Hence, x=3/2 is the only solution.