If \(2^a = 32\), \(b^4 = 81\), and \(b > 0\), then what is the value of \(a^b\)?
Solving for 'a'
We are given the equation \(2^a = 32\). To find the value of 'a', we need to express 32 as a power of 2.
We know that \(2 \times 2 \times 2 \times 2 \times 2 = 32\). Therefore, \(32 = 2^5\).
The equation becomes \(2^a = 2^5\).
By equating the exponents, we find that \(a = 5\).
Solving for 'b'
We are given the equation \(b^4 = 81\). To find 'b', we need to express 81 as a fourth power.
We know that \(3 \times 3 \times 3 \times 3 = 81\). Therefore, \(81 = 3^4\).
The equation becomes \(b^4 = 3^4\).
This gives two possible values for 'b': \(b = 3\) or \(b = -3\).
However, the problem states that \(b > 0\). So, we must choose \(b = 3\).
Calculating \(a^b\)
Now we need to find the value of \(a^b\) using the values we found for 'a' and 'b'.
We have \(a = 5\) and \(b = 3\).
Substitute these values into the expression \(a^b\): \(a^b = 5^3\).
Calculate the result: \(5^3 = 5 \times 5 \times 5 = 125\).
Thus, the value of \(a^b\) is 125.
Important Questions from Surds and Indices
The value of \(\frac{{{{\left( {251} \right)}^3} + {{\left( {249} \right)}^3}}}{{25.1 \times 25.1 - 624.99 + 24.9 \times 24.9}}\) is 5 × 10 k , where the value of k is :
Find the value of m in \(\left(\frac{2}{7}\right)^{-3} \times \left(\frac{2}{7}\right)^{-5}=\left (\frac{2}{7}\right)^{-3m+1}\)
If √625 = 25; then√(.00000625/25)is:
A. 0.0025
B. 0.001
C. 0.0001
D. 0.0005Find the value of:
\(\sqrt{150}-\sqrt{54}-\sqrt{24}\)
If \(\sqrt{4624}=68\) , then the value of:
\(\sqrt{46.24}+\sqrt{0.4624}+\sqrt{0.004624}\)
