Solve log base 2 of one sixteenth.

Hope this helps Rewrite

log

2

(

1

16

)

=

x

in exponential form using the definition of a logarithm. If

x

and

b

are positive real numbers and

b

does not equal

1

, then

log

b

(

x

)

=

y

is equivalent to

b

y

=

x

.

2

x

=

1

16

Create equivalent expressions in the equation that all have equal bases.

2

x

=

2

−

4

Since the bases are the same, the two expressions are only equal if the exponents are also equal.

x

=

−

4

The variable

x

is equal to  

−

4

.

−

4

Step-by-step explanation:

hello from mrbilldoesmath!

answer:   (7x + 6y) ( 49x^2 - 42xy + 36y^2)

discussion:

in general, the factorization of a^3 + b^3 is

a^3 + b^ 3 = (a + b) ( a^2 - ab + b^2)

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

(7x + 6y) ( 49x^2 - 42xy + 36y^2)

you,

mrb

(a + b) (a^2 - a b + b^2)