AI Tutor
What if you could take an equation with logarithms scattered on both sides and collapse them all into a single, straightforward quadratic?
In 1614, Scottish mathematician John Napier invented logarithms to turn tedious multiplication into simple addition, and that exact same principle allows us to merge multiple logarithmic terms today.
A logarithm log⁑a(y)=x\log_a(y) = x asks the question: "To what power must we raise base aa to get yy?", which means ax=ya^x = y with the restriction that base a>0,aβ‰ 1a > 0, a \neq 1 and the input argument y>0y > 0.
Because multiplying powers adds their exponents (auβ‹…av=au+va^u \cdot a^v = a^{u+v}), the log of a product splits into the sum of logs: log⁑a(M)+log⁑a(N)=log⁑a(MN)\log_a(M) + \log_a(N) = \log_a(MN).
How do we use this condensing power to crack an equation with unknown binomials inside the logs?

Condensing and Solving

Let's solve the equation log⁑a(x+3)+log⁑a(xβˆ’2)=log⁑a(6)\log_a(x + 3) + \log_a(x - 2) = \log_a(6) step by step.
First, we condense the left side into a single logarithm using the product law: log⁑a((x+3)(xβˆ’2))=log⁑a(6)\log_a((x + 3)(x - 2)) = \log_a(6).
By the One-to-One Property (injectivity) of logarithms, if log⁑a(A)=log⁑a(B)\log_a(A) = \log_a(B) (where a>0,aβ‰ 1a > 0, a \neq 1), then A=BA = B.
Expanding and rearranging gives x2+xβˆ’6=6x^2 + x - 6 = 6, which simplifies to the standard quadratic equation x2+xβˆ’12=0x^2 + x - 12 = 0.
Factoring gives (x+4)(xβˆ’3)=0(x + 4)(x - 3) = 0, yielding two potential algebraic solutions: x=3x = 3 and x=βˆ’4x = -4.