5.1: Ufafanuzi wa a
- Page ID
- 164577
Kwa idadi yoyote halisi\(a\) na idadi nzuri\(n\),\(a^n\) ni kuzidisha mara kwa mara ya\(a\) kwa\(n\) mara yenyewe.
\[a^n= a\cdot a \cdot a\cdot a\cdot a\cdot a\cdot a\cdot a \ldots \ldots \cdot a \nonumber \]
Nukuu:
\(a\)ni msingi,\(n\) ni exponent chanya.
\(a^n\)ni kusoma kama”\(a\) alimfufua kwa nguvu ya\(n\).”
Kutambua msingi na exponent katika maneno.
\(2^4\)\(x^5\),\(\left(\dfrac{3}{7}\right)^7\),\((-3)^3\)
Suluhisho
Ufafanuzi | Msingi | Mtetezi |
---|---|---|
\(2^4\) | 2 | 4 |
\(x^5\) | \(x\) | 5 |
\(\left(\dfrac{3}{7}\right)^7\) | \(\dfrac{3}{7}\) | 7 |
\((-3)^3\) | -3 | 3 |
Kutambua msingi na exponent ya yafuatayo.
Ufafanuzi | Msingi | Mtetezi |
---|---|---|
\(7^9\) | ||
\((-11)^6\) | ||
\(a^b\) | ||
\(\left(\dfrac{11}{12}\right)^5\) | ||
\(12^3\) | ||
\(\left(-\dfrac{7}{3}\right)^2\) | ||
\(x^7\) | ||
\((2.56)^4\) |
Kutathmini maneno ya fomu\(a^n\)
Wakati msingi na exponent ni thamani namba inawezekana kutathmini usemi imeandikwa katika na exponent. Ili kupata thamani, tumia ufafanuzi na kupanua maneno. Mara baada ya kupanua, kuzidisha na matokeo ni thamani ya namba ya kujieleza.
Panua maneno yafuatayo na tathmini ikiwa inawezekana.
\(3^4\),\(\left(\dfrac{3}{5}\right)^3\)\(x^7\),\((3.12)^2\),\((-5)^3\),\((-y)^6\)
Suluhisho
\(3^4\) | \(= 3\cdot 3\cdot 3\cdot 3 = 81\) |
\(\left(\dfrac{3}{5}\right)^3\) | \(\dfrac{3 }{5} \cdot \dfrac{3}{ 5 }\cdot \dfrac{3 }{5} = \dfrac{27 }{125}\) |
\(x^7\) |
\(x\cdot x\cdot x\cdot x\cdot x\cdot x\cdot x\) Kumbuka: Haiwezi kutathmini tangu x haijulikani |
\((3.12)^2\) | \((3.12)\cdot (3.12) = 9.734\) |
\((-5)^3\) | \(−5 \cdot −5 \cdot −5 = −12\) |
\((-y)^6\) |
\(−y \cdot −y \cdot −y \cdot −y \cdot −y \cdot −y = y^6\) Kumbuka: y haijulikani |
Panua maneno yafuatayo na tathmini ikiwa inawezekana.
- \(7^3\)
- \(\left(−\dfrac{ 2 }{3}\right)^4\)
- \((−x)^7\)
- \((7.14)^2\)
- \((−3)^9\)
- \((z)^5\)
- \(\left(− \dfrac{11 }{33 }\right)^2\)
- \(6^5\)
- \(\left(\dfrac{x}{ y}\right)^4\)
- \(a^{10}\)
- \(\left(\dfrac{2}{x}\right)^3\)