Answer :
Let's break down the given information and solve each part of the question step-by-step.
1. Given Information:
- [tex]\(\log_8(p) = 25 \)[/tex]
- [tex]\(\log_2(q) = 5 \)[/tex]
2. Finding [tex]\(p\)[/tex]:
- From [tex]\(\log_8(p) = 25\)[/tex]:
[tex]\[ p = 8^{25} \][/tex]
Simplifying [tex]\(8\)[/tex] as [tex]\(2^3\)[/tex], we get:
[tex]\[ p = (2^3)^{25} = 2^{75} \][/tex]
3. Finding [tex]\(q\)[/tex]:
- From [tex]\(\log_2(q) = 5\)[/tex]:
[tex]\[ q = 2^5 = 32 \][/tex]
4. Evaluating Expression for part (a) ([tex]\(p = q^{15}\)[/tex]):
- If [tex]\(p = q^{15}\)[/tex], we substitute [tex]\(q = 32\)[/tex]:
[tex]\[ q^{15} = 32^{15} \][/tex]
- Since [tex]\(32 = 2^5\)[/tex], we get:
[tex]\[ 32^{15} = (2^5)^{15} = 2^{75} \][/tex]
This shows that [tex]\(p = q^{15}\)[/tex].
Therefore, [tex]\(p = 2^{75} = 37778931862957161709568\)[/tex].
5. Evaluating Expression for part (b) ([tex]\(p\)[/tex]):
- From the given information, we already have [tex]\( p = 2^{75} \)[/tex].
Thus, [tex]\( p = 37778931862957161709568 \)[/tex].
6. Evaluating Expression for part (c) ([tex]\(p = q^5\)[/tex]):
- If [tex]\(p = q^5\)[/tex], we substitute [tex]\(q = 32\)[/tex]:
[tex]\[ q^5 = 32^5 \][/tex]
- Again, using [tex]\(32 = 2^5\)[/tex], we get:
[tex]\[ 32^5 = (2^5)^5 = 2^{25} \][/tex]
Thus, [tex]\(p = 2^{25} = 33554432\)[/tex].
### Summary of the Results:
- For part (a), [tex]\( p = 37778931862957161709568 \)[/tex]
- For part (b), [tex]\( p = 37778931862957161709568 \)[/tex]
- For part (c), [tex]\( p = 33554432 \)[/tex]
### Inequality:
Given inequality [tex]\( 4x + 3 < 6x + 7 \)[/tex]:
1. Start by isolating [tex]\( x \)[/tex]:
[tex]\[ 4x + 3 < 6x + 7 \][/tex]
2. Subtract [tex]\( 4x \)[/tex] from both sides:
[tex]\[ 3 < 2x + 7 \][/tex]
3. Subtract [tex]\( 7 \)[/tex] from both sides:
[tex]\[ -4 < 2x \][/tex]
4. Divide both sides by [tex]\( 2 \)[/tex]:
[tex]\[ -2 < x \][/tex]
Thus, [tex]\( x \)[/tex] belongs to the interval [tex]\( (-2, \infty) \)[/tex].
### Final Conclusion:
- The values of [tex]\(p\)[/tex] and [tex]\(q\)[/tex] are found.
- The interval for [tex]\(x\)[/tex] is [tex]\( (-2, \infty) \)[/tex].
1. Given Information:
- [tex]\(\log_8(p) = 25 \)[/tex]
- [tex]\(\log_2(q) = 5 \)[/tex]
2. Finding [tex]\(p\)[/tex]:
- From [tex]\(\log_8(p) = 25\)[/tex]:
[tex]\[ p = 8^{25} \][/tex]
Simplifying [tex]\(8\)[/tex] as [tex]\(2^3\)[/tex], we get:
[tex]\[ p = (2^3)^{25} = 2^{75} \][/tex]
3. Finding [tex]\(q\)[/tex]:
- From [tex]\(\log_2(q) = 5\)[/tex]:
[tex]\[ q = 2^5 = 32 \][/tex]
4. Evaluating Expression for part (a) ([tex]\(p = q^{15}\)[/tex]):
- If [tex]\(p = q^{15}\)[/tex], we substitute [tex]\(q = 32\)[/tex]:
[tex]\[ q^{15} = 32^{15} \][/tex]
- Since [tex]\(32 = 2^5\)[/tex], we get:
[tex]\[ 32^{15} = (2^5)^{15} = 2^{75} \][/tex]
This shows that [tex]\(p = q^{15}\)[/tex].
Therefore, [tex]\(p = 2^{75} = 37778931862957161709568\)[/tex].
5. Evaluating Expression for part (b) ([tex]\(p\)[/tex]):
- From the given information, we already have [tex]\( p = 2^{75} \)[/tex].
Thus, [tex]\( p = 37778931862957161709568 \)[/tex].
6. Evaluating Expression for part (c) ([tex]\(p = q^5\)[/tex]):
- If [tex]\(p = q^5\)[/tex], we substitute [tex]\(q = 32\)[/tex]:
[tex]\[ q^5 = 32^5 \][/tex]
- Again, using [tex]\(32 = 2^5\)[/tex], we get:
[tex]\[ 32^5 = (2^5)^5 = 2^{25} \][/tex]
Thus, [tex]\(p = 2^{25} = 33554432\)[/tex].
### Summary of the Results:
- For part (a), [tex]\( p = 37778931862957161709568 \)[/tex]
- For part (b), [tex]\( p = 37778931862957161709568 \)[/tex]
- For part (c), [tex]\( p = 33554432 \)[/tex]
### Inequality:
Given inequality [tex]\( 4x + 3 < 6x + 7 \)[/tex]:
1. Start by isolating [tex]\( x \)[/tex]:
[tex]\[ 4x + 3 < 6x + 7 \][/tex]
2. Subtract [tex]\( 4x \)[/tex] from both sides:
[tex]\[ 3 < 2x + 7 \][/tex]
3. Subtract [tex]\( 7 \)[/tex] from both sides:
[tex]\[ -4 < 2x \][/tex]
4. Divide both sides by [tex]\( 2 \)[/tex]:
[tex]\[ -2 < x \][/tex]
Thus, [tex]\( x \)[/tex] belongs to the interval [tex]\( (-2, \infty) \)[/tex].
### Final Conclusion:
- The values of [tex]\(p\)[/tex] and [tex]\(q\)[/tex] are found.
- The interval for [tex]\(x\)[/tex] is [tex]\( (-2, \infty) \)[/tex].