If log _{10}(x^{2}-6x+45)=2,then the values o | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Given the equation: $\log _{10}(x^{2}-6x+45)=2$By the definition of logarithm,$\log _{b}(a)=c$ implies $a=b^{c}$.Therefore,$x^{2}-6x+45=10^{2}$.$x

Vedclass · Accepted Answer

$11, -5$

Vedclass · Answer

(D) Given the equation: $\log _{10}(x^{2}-6x+45)=2$By the definition of logarithm,$\log _{b}(a)=c$ implies $a=b^{c}$.Therefore,$x^{2}-6x+45=10^{2}$.$x^{2}-6x+45=100$.Subtracting $100$ from both sides,we get $x^{2}-6x-55=0$.Factoring the quadratic equation: $(x-11)(x+5)=0$.Setting each factor to zero,we get $x-11=0$ or $x+5=0$.Thus,the values of $x$ are $x=11$ or $x=-5$.

If $\log _{10}(x^{2}-6x+45)=2$,then the values of $x$ are

Similar Questions

If the roots of the given equation $({m^2} + 1){x^2} + 2amx + {a^2} - {b^2} = 0$ are equal,then

If $5x + 9y = 5$ and $125x^3 + 729y^3 = 120$,then the value of the product of $x$ and $y$ is:

For the quadratic equation $2x^2 - (p + 1)x + (p - 1) = 0$,if $\alpha - \beta = \alpha \beta$,then what is the value of $p$?

The roots of the equation $x^{2/3} + x^{1/3} - 2 = 0$ are

Solve the given two equations and choose the correct option.
$I.$ $2x^{2} - 13x + 21 = 0$
$II.$ $5y^{2} - 22y + 21 = 0$

Vedclass Test Series

Exam Paper Generator

Online Exam Module