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2 years ago

Solving Exponential equation 7^7x+5 = 2^x-1

Mathematics

1 answer:

JulijaS [17]2 years ago

4 0

Answer:

x=-6/7^7-2

Step-by-step explanation:

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2 years ago

Use the rules of exponents to simplify the expressions. Match the expression with its equivalent value.

Lelechka [254]

Answer:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}} = 32$

Step-by-step explanation:

1) $\frac{(-2)^{-5}}{(-2)^{-10}}$

Solving using exponent rule: $a^{-m}=\frac{1}{a^m}$

$\frac{(-2)^{-5}}{(-2)^{-10}}\\=(-2)^{-5+10}\\=(-2)^{5}\\=-32$

So, $\frac{(-2)^{-5}}{(-2)^{-10}}=-32$

2) $2^{-1}.2^{-4}$

Using the exponent rule: $a^m.a^n=a^{m+n}$

We have:

$2^{-1}.2^{-4}\\=2^{-1-4}\\=2^{-5}$

We also know that: $a^{-m}=\frac{1}{a^m}$

Using this rule:

$2^{-5}\\=\frac{1}{2^5}\\=\frac{1}{32}$

So, $2^{-1}.2^{-4} = \frac{1}{32}$

3) $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2$

Solving:

$(-\frac{1}{2} )^3.(-\frac{1}{2} )^2\\=(-\frac{1}{8} ).(\frac{1}{4} )\\=-\frac{1}{32}$

So, $(-\frac{1}{2} )^3.(-\frac{1}{2} )^2=-\frac{1}{32}$

4) $\frac{2}{2^{-4}}$

We know that: $a^{-m}=\frac{1}{a^m}$

$\frac{2}{2^{-4}}\\=2\times 2^4\\=2(16)\\=32$

So, $\frac{2}{2^{-4}} = 32$

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2 years ago