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Solve for [tex]\( x \)[/tex]:
[tex]\[ \sqrt{7} \div 49^x = \frac{1}{7} \][/tex]


Sagot :

Let's solve the given equation step-by-step:

[tex]\[ \sqrt{7} \div 49^x = \frac{1}{7} \][/tex]

First, we need to express the terms involving roots and powers in a common base. We know:

[tex]\[ \sqrt{7} = 7^{1/2} \][/tex]

Also, we can express 49 as a power of 7:

[tex]\[ 49 = 7^2 \quad \text{so} \quad 49^x = (7^2)^x = 7^{2x} \][/tex]

Now, substituting these into the original equation, we get:

[tex]\[ 7^{1/2} \div 7^{2x} = \frac{1}{7} \][/tex]

Using properties of exponents, specifically [tex]\(a^m \div a^n = a^{m-n}\)[/tex], we can simplify the left-hand side:

[tex]\[ 7^{1/2 - 2x} = \frac{1}{7} \][/tex]

Next, we rewrite [tex]\(\frac{1}{7}\)[/tex] as a power of 7. Since [tex]\(\frac{1}{7} = 7^{-1}\)[/tex], we now have:

[tex]\[ 7^{1/2 - 2x} = 7^{-1} \][/tex]

Since the bases (7) are the same on both sides of the equation, we can set the exponents equal to each other:

[tex]\[ \frac{1}{2} - 2x = -1 \][/tex]

To isolate [tex]\(x\)[/tex], we solve the equation step-by-step:

1. Add 1 to both sides:

[tex]\[ \frac{1}{2} - 2x + 1 = -1 + 1 \][/tex]

[tex]\[ \frac{1}{2} + 1 - 2x = 0 \][/tex]

2. Combine the constants on the left side:

[tex]\[ \frac{1}{2} + \frac{2}{2} - 2x = 0 \][/tex]

[tex]\[ \frac{3}{2} - 2x = 0 \][/tex]

3. Add [tex]\(2x\)[/tex]:

[tex]\[ \frac{3}{2} = 2x \][/tex]

4. Divide both sides by 2 to solve for [tex]\(x\)[/tex]:

[tex]\[ x = \frac{3}{2} \div 2 = \frac{3}{4} \][/tex]

Thus, we find:

[tex]\[ x = \frac{3}{4} \][/tex]

The step-by-step solution demonstrates that the value of [tex]\(x\)[/tex] that satisfies the given equation [tex]\(\sqrt{7} \div 49^x = \frac{1}{7}\)[/tex] is:

[tex]\[ x = \frac{3}{4} \][/tex]