All categories

3 years ago

Log4(x+2)-log4(x+3)=-1

Mathematics

1 answer:

vova2212 [387]3 years ago

4 0

Answer:

Step-by-step explanation:

I take it that the base of the log is 4.

So the question can be written as

log_4 [(x+2)/(x + 3)] = -1

If you take the antilog of this, you get

(x + 2)/(x+3) = 4^-1 = 1/4

Now Cross Multiply

4*(x + 2) = 1*(x + 3) Remove the brackets on both sides

4x + 8 = x + 3 subtract x from both sides.

4x - x + 8 = x - x + 3 Combine

3x + 8 = 3 Subtract 8 from both sides

3x = 3 - 8 Combine

3x = - 5 Divide by 3

x = - 5/3

You might be interested in

|−9| − |−5| i hate next gen but help pls

Setler79 [48]

Answer:

Step-by-step explanation:

$\mathrm{-9 = 9, -5 = 5,\:because\:they\:are\:in\:absolute\:value}\\9 - 5 = 4$

8 0

1 year ago

Read 2 more answers

A circle has a circumference of 44m. What

MakcuM [25]

<h3>Answer: B. 154 square meters</h3>

================================

Work Shown:

C = 44 is the circumference

in general, the circumference of a circle is

C = 2*pi*r

Plug in C = 44 and solve for r.

C = 2*pi*r

44 = 2*pi*r

2*pi*r = 44

r = 44/(2pi)

r = 7.0028175 which is approximate

--------------------

Now plug this into the formula for the area of a circle

A = pi*r^2

A = pi*(7.0028175)^2

A = 154.061985

A = 154 rounding to the nearest whole number

3 0

3 years ago

Read 2 more answers

#82 will give brainliest to best answer!

Fofino [41]

Answer:

$\frac{6}{k-6}$

Step-by-step explanation:

First, we can factor all of the following equations to turn that weird, huge looking thing into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ ÷ $\frac{(k-6)^2}{k(k-6)}$ × $\frac{6(k-10)}{k(k + 6)}$ . We know that division is simply multiplication by the reciprocal, so that whole equation will turn into $\frac{(k+6)(k-6)}{(k-6)(k-10)}$ × $\frac{k(k-6)}{(k-6)^2}$ × $\frac{6(k-10)}{k(k+6)}$ . Now we can cancel out some values if they are both in the numerator and denominator, which will turn that still huge looking thing into $\frac{6}{k-6}$ which is our final answer, as it cannot be simplified further.