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4 years ago

Solve the exponential equation for x 216=6^(4x+11)

Mathematics

2 answers:

olya-2409 [2.1K]4 years ago

8 0

Answer:

x=-2

Step-by-step explanation:

216=6^(4x+11)

$216=6^{4x+11}$

To solve for x , we need to make same base on both sides

Write 216 in exponential form

216= 6 times 6 times 6 = 6^3

$6^3=6^{4x+11}$

Both sides have same base 6. So we make the exponents equal and solve for x

$3=4x+11$

subtract 11 on both sides

$-8=4x$

Divide both sides by 4

$-2=x$

Anastasy [175]4 years ago

3 0

The answer is negatif 2

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Step-by-step explanation:

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Step-by-step explanation:

2 2/3 × 1 5/7

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3 years ago

What is (8-√2)(2+√8)?

siniylev [52]

Answer:

$(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = 12 + 14\; \sqrt{2}$ .

Step-by-step explanation:

Step One: Simplify the square roots.

$8 = 4 \times 2 = 2^2 \times 2$ .

The square root of 8 $\sqrt{8}$ can be simplified as the product of an integer and the square root of 2:

$\sqrt{8} = \sqrt{4 \times 2} = \sqrt{2^2 \times 2} = \sqrt{2^{2}} \times \sqrt{2} = 2 \; \sqrt{2}$ .

As a result,

$(8 - \sqrt{2}) \cdot (2 + \sqrt{8}) = (8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})$ .

Step Two: Expand the product of the two binomials.

$(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})$ is the product of two binomials:

Binomial One: $8 - \sqrt{2}$ .
Binomial Two: $2 + 2\; \sqrt{2}$

Start by applying the distributive law to the first binomial. Multiply each term in the first binomial (without brackets) with the second binomial (with brackets)

$({\bf 8} - {\bf \sqrt{2}}) \cdot {(2 + 2\; \sqrt{2})}\\= [{\bf 8} \cdot {(2 + 2\; \sqrt{2})}] - [{\bf \sqrt{2}} \cdot {(2 + 2\; \sqrt{2})}]$

Now, apply the distributive law once again to terms in the second binomial.

$[8 \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})] - [\sqrt{2} \cdot ({\bf 2} + {\bf 2\; \sqrt{2}})]\\= [8 \times {\bf 2} + 8 \times {\bf 2\;\sqrt{2}}] - [\sqrt{2} \times {\bf 2} + \sqrt{2} \times {\bf 2\; \sqrt{2}}]$ .

Step Three: Simplify the expression.

The square of a square root is the same as the number under the square root. For example, $\sqrt{2} \times \sqrt{2} = (\sqrt{2})^{2} = 2$ .

$[8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + 2 \sqrt{2} \times \sqrt{2}]\\ =[16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4$ .

Combine the terms with the square root of two and those without the square root of two:

$16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})$ .

Factor the square root of two out of the second term:

$(16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\= (16 - 4) + (16 - 2) \; \sqrt{2} \\= 12 - 14 \; \sqrt{2}$ .

Combining the steps:

$(8 - \sqrt{2}) \cdot (2 + 2\; \sqrt{2})\\= [8 \cdot (2 + 2\; \sqrt{2})] - [\sqrt{2} \cdot (2 + 2\; \sqrt{2})]\\= [8 \times 2 + 8 \times 2\;\sqrt{2}] - [\sqrt{2} \times 2 + \sqrt{2} \times 2 \;\sqrt{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times (\sqrt{2})^{2}]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 2 \times 2]\\= [16 + 16 \;\sqrt{2}] - [2 \;\sqrt{2} + 4]\\= 16 + 16\;\sqrt{2} - 2\; \sqrt{2} - 4\\= (16 - 4) + (16 \; \sqrt{2} - 2\; \sqrt{2})\\$

$= (16 - 4) + (16 - 2) \; \sqrt{2}\\= 12 - 14 \; \sqrt{2}$ .

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