Question: Solve log5x = 4 logx5
What I have done: Is it correct?
log5x = 4 / log5x
( log5x)2= 4
log5x = 2 or log5x = -2
x = 52 x = 5 - 2 x = 25 x = 1/25 = 0.04
I have found 2 answers.
But answer given is 25 . Why only one answer?
Can x = 0.04 be accepted as one of the answers?
Does this equation has only 1 solution or 2 solutions?
In your first step you are using the fact that \(\displaystyle log_a(b)= \frac{1}{log_b(a)}\). That is true because if \(\displaystyle log_a(b)= X\) then \(\displaystyle b= a^X\). Solve for a by taking the 1/X power of both sides: \(\displaystyle b^{1/X}= a\) So that \(\displaystyle log_b(a)= \frac{1}{X}= \frac{1}{log_a(b)}\).
Now, in general, when you multiply both sides of an equation by something involving the unknown you
might introduce "extraneous solutions", solutions to the
new equation that do not satisfy the original equation. As a very simple example, x- 2= 0 has the obvious solution x= 2. If I multiply both sides of the equation by x- 3 I get [/tex](x- 2)(x- 3)= x^2- 5x+ 6= 0[/tex] which has solutions x= 2 and x= 3. In this problem you introduced the "extraneous solution" x= 1/25 when you multiplied both sides of the equation by \(\displaystyle log_5(x)\). That is why, even when you are sure you have done everything correctly, you should always check your answers in the original equation:
If \(\displaystyle x= 25= 5^2\), so that \(\displaystyle 5= 25^{1/2}\), then \(\displaystyle log_5(25)= 2\) while \(\displaystyle 4 log_{25}(5)= 4\left(\frac{1}{2}\right)= 2\). Yes, \(\displaystyle 2= 4\left(\frac{1}{2}\right)\).
But if \(\displaystyle x= \frac{1}{25}= 5^{-2}\) then \(\displaystyle log_5(1/25)= -2\) while \(\displaystyle 5= \frac{1}{25^{1/2}}= 25^{-1/2}\) so \(\displaystyle log_{1/25}(5)= -\frac{1}{2}\). It
is true that \(\displaystyle log_5(1/25)= 4 log_{1/25}(5)\) so that
NO, X= 1/25 IS NOT AN "EXTRANEOUS" SOLUTION IT IS A PERFECTL VALID SOLUTION!
