1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
uranmaximum [27]
3 years ago
10

Solve (1/81)^x*1/243=(1/9)^−3−1 by rewriting each side with a common base.

Mathematics
1 answer:
elena55 [62]3 years ago
8 0

Answer:

x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]

Step-by-step explanation:

you have the following formula:

(\frac{1}{81})^{\frac{x}{243}}=(\frac{1}{9})^{-3}-1

To solve this equation you use the following properties:

log_aa^x=x

Thne, by using this propwerty in the equation (1) you obtain for x

log_{(\frac{1}{81})}(\frac{1}{81})^{\frac{x}{243}}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\\frac{x}{243}=log_{\frac{1}{81}}[(\frac{1}{81})-1]\\\\x=(243)log_{\frac{1}{81}}[(\frac{1}{81})-1]

You might be interested in
Help.........................<br><br>​
garri49 [273]
<h3>Answer: Choice A</h3>

x^2\left(\sqrt[4]{x^2}\right)

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

Explanation:

The fourth root of x is the same as x^(1/4)

I.e,

\sqrt[4]{x} = x^{1/4}

The same applies to x^10 as well

\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}

Multiply the exponents 10 and 1/4 to get 10/4

\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}

\sqrt[4]{x^{10}} = x^{10/4}

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

If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below

x^{m/n} = x^a\sqrt[n]{x^b}

The 'a' and 'b' are found through dividing m/n

m/n = a remainder b

'a' is the quotient, b is the remainder

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

The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4

m/n = 10/4 = 2 remainder 2

We have a = 2 and b = 2

So

x^{m/n} = x^a\sqrt[n]{x^b}

turns into

x^{10/4} = x^2\sqrt[4]{x^2}

which means

\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}

7 0
3 years ago
Write the expression that represents length in terms of width for this problem
raketka [301]
Assuming you are talking about the rectangle, 

width = area/lenght
5 0
3 years ago
16 - 2t = 5t + 9<br><br> What number does t represent? I'm bored
seraphim [82]
Try not to be bored.  You can take the attitude that you actually like math, and that you will some day actually be a MathGenius.

t represents an unknown quantity defined by the equation 16 - 2t = 5t + 9.

To find this value, do the following:  Add 2t to both sides, and then subtract 9 from both sides.  You'll end up with 7t = 7, which yields t = 1.

We call this the "solution" of the equation given.
8 0
3 years ago
Read 2 more answers
A kite in the shape of a square with a diagonal 32 cm and an isosceles triangle of base 8 cm and sides 6 cm each is to be made o
alexgriva [62]

Answer:

this question is from which class

Step-by-step explanation:

<em>what's your class </em>

5 0
4 years ago
Read 2 more answers
Brainers, please help me with this problem.
Elenna [48]

Answer:

Step-by-step explanation:

12x^2-30?

8 0
3 years ago
Read 2 more answers
Other questions:
  • the coordinates of the endpoints of CE are C(-3, -7) and E(7, -2). point M (3, -4) is on CE. what is the ratio of CM:ME?
    15·1 answer
  • Which type of angle is angle ABC
    7·1 answer
  • D) (K<br> If f(x) = x3 and g(x) = 2x + 7, what is g(x)<br> when x = 2?
    9·1 answer
  • Helpppppppppppppppppppppppppppp
    10·2 answers
  • 1/4x - 8=4. I am not understanding this question can someone help me plssss.
    12·2 answers
  • Y&gt; 3x-2<br> Graph the solution
    9·1 answer
  • Please help!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
    14·2 answers
  • Help me PLEASEEEEEEEEEEEEE
    12·1 answer
  • a linear function contains the following points. what are the slope and y-intercept of this function?
    14·1 answer
  • Which shapes make up this composite figure?
    10·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!