if 3^x = 12 then 3^(2+x) =?

Question

another question: ( 6x^2+16x+10 ) / (x^3 - 4x^2) =0 then x =? plz show ur workings my aim is not th answer rather how to solve

Answer 1

If 3^x = 12, then 3^(2 + x) = ?

Using the exponent property b^(x + y) = b^x b^y,
3^(2 + x) = 3² · 3^x

Substitute 3^x
3^(2 + x) = 9 · 12

Multiply
3^(2 + x) = 108

another answer:
(6x² + 16x + 10) / (x³ - 4x²) = 0

Factorize both the numerator and the denominator:
2(3x + 5)(x + 1)/x²(x - 4) = 0

Multiply x²(x - 4) to both sides (or simply criss cross)
2(3x + 5)(x + 1) = 0

Divide the equation by 2
(3x + 5)(x + 1) = 0

Thus,
3x + 5 = 0 or x + 1 = 0

Therefore,
x = -5/3 or x = -1

^_^

Answer 2

3^(2+x)
= 3^2 * 3^x(using a^(b+c) = a^b*a^c)
= 9* 12 = 108 (3^2 = 9 and 3^x = 12 given
(6x^2+16x+10)/(x^3-4x^2)
numeratior = 6x^2+ 6x + 10x +10
= 6x(x+1) + 10(x+1) = (6x+10)(x+1) = 2(3x+5)(x+1)
denominator = x^3-4x^2 = x(x-4)
we have to ensure that numerator = 0 and denominator !=0
2(x+1)(3x+5) = 0 ; there is no common factor
x+1 = 0 or 3x+5 = 0
x = -1 or -5/3

Answer 3

3^(2+x)=3^2*3^x=9*12=108
6x^2+16x+10=6x^2+6x+10x+10=(6x+10)(x+1)
x^3-4x^2=x^2(x-4)
if Nr/Dr=0 then Nr =-0
(6x+10)(x+1)=0
6x+10=0 or x=-5/3 and x+1=0 or x=-1
(and by the way x cannot be 4 as then the denominator will reduce to zero)

Answer 4

a million. what's the fringe of a sq. with the size of the two aspects being 8 inches? 8+8+8+8= 32 inches 2.what's the area of a rectangle with a width of five inches and a length of 9 inches. 5x9= fifty 4 inches squared 3.what's the fringe of a rectangle with a width of 6 cm and a length of 10 cm? 6+6+10+10= 32 cm 4.what's the area of a sq. with the size of the two aspects being 8 cm? 8x8=sixty 4 cm squared 5.what's the area of a triangle with a base of three cm and a top of 12 cm? uncertain 6.what's the area of a parallelogram with a base of 6 cm and a top of 9 cm? uncertain 7.what's the area of a trapezoid with a top of five inches and bases of b1 = 8 inches and b2 = 6 inches? forget the equation 8.what's the answer set for 5 – 2x = a million if the replace set is {-2, -a million, 0, a million, 2}? 9.remedy: x – 5 = -3 x + -5=-3 x=-3+-5 x=-8 10.remedy: x + 5 = 9. x-5=9 x=9-5 x=4 11.remedy: x_ = -2 6 uncertain of the concern............ 12.remedy: -3x = 12 x/-3=12 x=12/-3 x=-4 13.remedy: 2x – a million = 7 2x+3=7 2x=7+3 2x=10 x=10/2 x=5 14.remedy: a million x + 6 = 9 x+6=9 x=9-6 x=3 15.remedy -5x + 3 = 23 -5x+3-3=23-3 -5x=20 x=20/-5 x=-4 sixteen.4x -2 = -14 4x-2+2=-14+2 4x=-12 x=-12/4 x=-3

Answer 5

3^(2+x) =3^x*3^2=12*9=108
( 6x^2+16x+10 ) / (x^3 - 4x^2) =0
it only exist when the denominator is not 0
now the numerator is ( 6x^2+16x+10 )
=(6x+10)*(x+1)
equaqting it to 0
6x+10=0 i.e. x=-5/3 & for x+1=0: x=-1
now the denominator is -15.74 for x=-5/3 and -5 for x=-1. so denominator is not 0

Answer 6

3^x=12 => x=log 3 (12)=1+log3 (4) { log base is 3}
so 3^(2+x)= 3^[3+log3 (4)] = 108


(6x^2+16x+10)/(x^3-4x^2)=0 implies x is not equal to 0 or 4
We also get
6x^2+16x+10=0 so x=-1 or x=-5/3

Answer 7

3^x = 12, then x= log12base 3.

log12base3= log 4x3 base3

so log 12base3= log4base3 + log3base3

since log3base3=1

log12base3= 1+ log4base3

3^ (2+x) = 3^ (2+ 1+ log4base3)

3^(2+x) = 3^ ( 3+ log4base3 )

3^ ( 3+ log4base3)= 3^3 + 3^log4base

by equation, we know that 3^log4base3 = 4^log3base3

log3base3 is equal to 1.

so we get 3^(2+x) = 3^3 + 4^log3base3

3^ (2+x) = 27+ 4^1

3^(2+x) = 27+ 4

3^ ( 2+ x) = 31

so if 3^x = 12, we get that 3^(2+x) = 31.

hope u understand

Answer 8

3^(2+x)=(3^2)*(3^x)=12*9
because of that the fraction is zero then the expresion 6x^2+16x+10 should equals zero 6x^2+16x+10=(x+1)(6x+10)
then x=-1 or x=-10/6

Answer 9

3^x=12
Take log base 3 of both sides

log3 12 = x

plug into other eq

=3^(2+log3 12)
=3^(2)*3^(log3(12))
The trick is at this step where anything taken to a power that's has a log of the same base in the exponent can be simplified to be the number that your taking the log of.
=3^2*12
=9*12=108

Answer 10

Given,

3^x = 12

Multiply both sides by 3^2,

3^x (3^2) = 12 (3^2)

3^(2 + x) = 12 (9)

So, 3^(2 + x) = 108