Answer :

Let's solve each of the equations step-by-step:

### i. [tex]\( 2^x = 10 \)[/tex]

1. Take the natural logarithm of both sides:
[tex]\[ \ln(2^x) = \ln(10) \][/tex]

2. Utilize the property of logarithms: [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ x \cdot \ln(2) = \ln(10) \][/tex]

3. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\ln(10)}{\ln(2)} \][/tex]

4. Approximate the values of the logarithms:
[tex]\[ \ln(10) \approx 2.302585 \][/tex]
[tex]\[ \ln(2) \approx 0.693147 \][/tex]

5. Perform the division:
[tex]\[ x \approx \frac{2.302585}{0.693147} \approx 3.321928 \][/tex]

So, the solution to the first equation is:
[tex]\[ x \approx 3.321928 \][/tex]

### ii. [tex]\( 5^{4x-1} = 7^{x+1} \)[/tex]

1. Take the natural logarithm of both sides:
[tex]\[ \ln(5^{4x-1}) = \ln(7^{x+1}) \][/tex]

2. Utilize the property of logarithms: [tex]\(\ln(a^b) = b \ln(a)\)[/tex]:
[tex]\[ (4x - 1) \cdot \ln(5) = (x + 1) \cdot \ln(7) \][/tex]

3. Distribute the logarithms:
[tex]\[ 4x \cdot \ln(5) - \ln(5) = x \cdot \ln(7) + \ln(7) \][/tex]

4. Rearrange to isolate terms involving [tex]\(x\)[/tex]:
[tex]\[ 4x \cdot \ln(5) - x \cdot \ln(7) = \ln(7) + \ln(5) \][/tex]

5. Factor out [tex]\(x\)[/tex] on the left-hand side:
[tex]\[ x \cdot (4 \ln(5) - \ln(7)) = \ln(7) + \ln(5) \][/tex]

6. Solve for [tex]\(x\)[/tex]:
[tex]\[ x = \frac{\ln(7) + \ln(5)}{4 \ln(5) - \ln(7)} \][/tex]

7. Approximate the values of the logarithms:
[tex]\[ \ln(5) \approx 1.609438 \][/tex]
[tex]\[ \ln(7) \approx 1.945910 \][/tex]

8. Substitute the approximate values:
[tex]\[ x \approx \frac{1.945910 + 1.609438}{4 \cdot 1.609438 - 1.945910} \][/tex]

9. Perform the arithmetic:
[tex]\[ x \approx \frac{3.555348}{4 \cdot 1.609438 - 1.945910} \approx \frac{3.555348}{6.437752 - 1.945910} \approx \frac{3.555348}{4.491842} \approx 0.791512 \][/tex]

So, the solution to the second equation is:
[tex]\[ x \approx 0.791512 \][/tex]

### Summary
The solutions to the given equations are approximately:
i. [tex]\( x \approx 3.321928 \)[/tex]
ii. [tex]\( x \approx 0.791512 \)[/tex]