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BabaBlast [244]
3 years ago
5

X to the power of negative 10 times x to the power 6

Mathematics
1 answer:
Llana [10]3 years ago
6 0
X^-10×x^6
=x^-4
=1/x^4

Hope my answer helped u :)
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What is the equation of log x 729=3?
pishuonlain [190]

Answer:

Step-by-step explanation:

Move 729  to the left side of the equation by subtracting it from both sides.  x  3 − 729  = 0  Factor the left side of the equation.  Rewrite  729  as 9

3

.  x

3

−

9

3

=

0

. Since both terms are perfect cubes, factor using the difference of cubes formula,  a

3

−

b

3

=

(

a

−

b

)

(

a

2+ab+b2). Where a

=x and  b=9. (x−9)(x2+x⋅9+92)=0

. Simplify.  Move 9  to the left of x

.  (x−9)(x2+9x+92)=0.  Raise  9  to the power of 2

.  (x

−9

)(

x

2

+

9

x

+81

)=0

. Set x

−9  equal to  0 and solve for x.  Set the factor equal to  0. x−

9=

0. Add 9  to both sides of the equation.  x=9

. Set x2+

9

x

+

81 equal to  0  and solve for  x

.  Set the factor equal to 0

.  x2+9x+81=0.  Use the quadratic formula to find the solutions.  −b±√b2−4(ac) 2a. Substitute the values  a=1, b=9, and c=81  into the quadratic formula and solve for  x.  −9±√92−4⋅ (1⋅81

) 2⋅

1  Simplify.  Simplify the numerator. Raise 9  to the power of 2. x=−9±√81−4⋅(1⋅81) 2⋅1. Multiply  

81

by  

1

.

x

=

−

9

±

√

81

−

4

⋅

81

2

⋅

1

Multiply  

−

4

by  

81

.

x

=

−

9

±

√

81

−

324

2

⋅

1

Subtract  

324

from  

81

.

x

=

−

9

±

√

−

243

2

⋅

1

Rewrite  

−

243

as  

−

1

(

243

)

.

x

=

−

9

±

√

−

1

⋅

243

2

⋅

1

Rewrite  

√

−

1

(

243

)

as  

√

−

1

⋅

√

243

.

x

=

−

9

±

√

−

1

⋅

√

243

2

⋅

1

Rewrite  

√

−

1

as  

i

.

x

=

−

9

±

i

⋅

√

243

2

⋅

1

Rewrite  

243

as  

9

2

⋅

3

.

Tap for fewer steps...

Factor  

81

out of  

243

.

x

=

−

9

±

i

⋅

√

81

(

3

)

2

⋅

1

Rewrite  

81

as  

9

2

.

x

=

−

9

±

i

⋅

√

9

2

⋅

3

2

⋅

1

Pull terms out from under the radical.

x

=

−

9

±

i

⋅

(

9

√

3

)

2

⋅

1

Move  

9

to the left of  

i

.

x

=

−

9

±

9

i

√

3

2

⋅

1

Multiply  

2

by  

1

.

x

=

−

9

±

9

i

√

3

2

Factor  

−

1

out of  

−

9

±

9

i

√

3

.

x

=

−

1

9

±

9

i

√

3

2

Multiply  

−

1

by  

−

1

.

x

=

1

−

9

±

9

i

√

3

2

Multiply  

−

9

±

9

i

√

3

by  

1

.

x

=

−

9

±

9

i

√

3

2

The final answer is the combination of both solutions.

x

=

−

9

−

9

i

√

3

2

,

−

9

+

9

i

√

3

2

The solution is the result of  

x

−

9

=

0

and  

x

2

+

9

x

+

81

=

0

.

x

=

9

,

−

9

−

9

i

√

3

2

,

−

9

+

i

√

3

2

7 0
3 years ago
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