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Question: The set of solutions for the equation \( {\log _{10}}\left( {{a^2} - 15a} \right) = 2 \) consists of...

The set of solutions for the equation log⁡10(a2−15a)=2{\log _{10}}\left( {{a^2} - 15a} \right) = 2 consists of?
A) Two integers
B) One integer and one fraction
C) Two irrational numbers
D) Two non-real numbers
E) No numbers, that is, set is empty

Explanation

Solution

Hint : We can solve the equation by opening the log and then forming the quadratic equation. Then we can find the solutions to quadratic equations by factorizing or by using the quadratic formula.
Formula used-
Formulae used are
log⁡ay=x⇔ax=y{\log _a}y = x \Leftrightarrow {a^x} = y where a≠0,1,y≠0a \ne 0,1,y \ne 0
x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} for a quadratic equation ax2+bx+c=0a{x^2} + bx + c = 0 where a≠0a \ne 0

Complete step-by-step answer :
According to the question,
log⁡10(a2−15a)=2{\log _{10}}\left( {{a^2} - 15a} \right) = 2
We can remove the log and convert the equation to simple algebra by
using, log⁡ay=x⇔ax=y{\log _a}y = x \Leftrightarrow {a^x} = y we can see
a2−15a=102 ⇒a2−15a−100=0  {a^2} - 15a = {10^2} \\\ \Rightarrow {a^2} - 15a - 100 = 0 \\\
Now that we have got a quadratic equation. We can find the roots by any method.
Using the quadratic formula, x=−b±b2−4ac2ax = \dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}
a=−(−15)±(−15)2−4×1×(−100)2.1 ⇒a=15±225+4002 ⇒a=15±6252=15±252  ⇒a=20,−5   a = \dfrac{{ - \left( { - 15} \right) \pm \sqrt {{{\left( { - 15} \right)}^2} - 4\times1\times\left( { - 100} \right)} }}{{2.1}} \\\ \Rightarrow a = \dfrac{{15 \pm \sqrt {225 + 400} }}{2} \\\ \Rightarrow a = \dfrac{{15 \pm \sqrt {625} }}{2} = \dfrac{{15 \pm 25}}{2} \; \Rightarrow a = 20, - 5 \;
So here we can see that the solution set for the initial equation is (20,−5)(20,-5) and the set consists of two integers. Therefore, the correct answer is A) Two integers.
So, the correct answer is “Option A”.

Note : We can also find the roots using factorization as following
a2−15a−100=0 ⇒a2−20a+5a−100=0 ⇒a(a−20)+5(a−20)=0 ⇒(a−20)(a+5)=0 ⇒a=20,−5   {a^2} - 15a - 100 = 0 \\\ \Rightarrow {a^2} - 20a + 5a - 100 = 0 \\\ \Rightarrow a\left( {a - 20} \right) + 5\left( {a - 20} \right) = 0 \\\ \Rightarrow \left( {a - 20} \right)\left( {a + 5} \right) = 0 \\\ \Rightarrow a = 20, - 5 \;