Answer :
Sure! Let's solve each part of the question step-by-step:
a) [tex]\(\log_2 x = 5\)[/tex]
To find [tex]\(x\)[/tex], we use the property of logarithms that says [tex]\(x = \text{base} ^ \text{exponent}\)[/tex]. Therefore, [tex]\(x = 2^5\)[/tex].
So, [tex]\(x = 32\)[/tex].
b) [tex]\(\log_4 16 = x\)[/tex]
We need to express 16 as a power of 4. Notice that [tex]\(16 = 4^2\)[/tex].
So, [tex]\(x = 2\)[/tex].
c) [tex]\(\log_7 7 = x\)[/tex]
Any number raised to the power of 1 is itself, so [tex]\(x = 1\)[/tex].
d) [tex]\(\log_x 16 = 2\)[/tex]
This can be rewritten using the definition of logarithms: [tex]\(x^2 = 16\)[/tex].
Taking the square root of both sides, [tex]\(x = \sqrt{16}\)[/tex].
So, [tex]\(x = 4\)[/tex].
e) [tex]\(\log_8 x = \frac{1}{3}\)[/tex]
This means [tex]\(8^{\frac{1}{3}} = x\)[/tex]. Recognizing that [tex]\(8\)[/tex] is [tex]\(2^3\)[/tex], we see:
[tex]\[ x = (2^3)^{\frac{1}{3}} = 2^{3 \times \frac{1}{3}} = 2^1 = 2 \][/tex]
So, [tex]\(x = 2\)[/tex].
f) [tex]\(\log_{\frac{1}{3}} 9 = x\)[/tex]
Using the property [tex]\(b^y = z\)[/tex] for logarithms, we have [tex]\((\frac{1}{3})^x = 9\)[/tex].
Recognizing [tex]\(9\)[/tex] as [tex]\(3^2\)[/tex], we rewrite the equation:
[tex]\((3^{-1})^x = 3^2 \)[/tex], which simplifies to [tex]\(3^{-x} = 3^2\)[/tex].
Equating the exponents, [tex]\(-x = 2\)[/tex], so [tex]\(x = -2\)[/tex].
So [tex]\(x = -2\)[/tex].
These steps will lead us to the answers:
- a) [tex]\(32\)[/tex]
- b) [tex]\(2\)[/tex]
- c) [tex]\(1\)[/tex]
- d) [tex]\(4\)[/tex]
- e) [tex]\(2\)[/tex]
- f) [tex]\(-2\)[/tex]
a) [tex]\(\log_2 x = 5\)[/tex]
To find [tex]\(x\)[/tex], we use the property of logarithms that says [tex]\(x = \text{base} ^ \text{exponent}\)[/tex]. Therefore, [tex]\(x = 2^5\)[/tex].
So, [tex]\(x = 32\)[/tex].
b) [tex]\(\log_4 16 = x\)[/tex]
We need to express 16 as a power of 4. Notice that [tex]\(16 = 4^2\)[/tex].
So, [tex]\(x = 2\)[/tex].
c) [tex]\(\log_7 7 = x\)[/tex]
Any number raised to the power of 1 is itself, so [tex]\(x = 1\)[/tex].
d) [tex]\(\log_x 16 = 2\)[/tex]
This can be rewritten using the definition of logarithms: [tex]\(x^2 = 16\)[/tex].
Taking the square root of both sides, [tex]\(x = \sqrt{16}\)[/tex].
So, [tex]\(x = 4\)[/tex].
e) [tex]\(\log_8 x = \frac{1}{3}\)[/tex]
This means [tex]\(8^{\frac{1}{3}} = x\)[/tex]. Recognizing that [tex]\(8\)[/tex] is [tex]\(2^3\)[/tex], we see:
[tex]\[ x = (2^3)^{\frac{1}{3}} = 2^{3 \times \frac{1}{3}} = 2^1 = 2 \][/tex]
So, [tex]\(x = 2\)[/tex].
f) [tex]\(\log_{\frac{1}{3}} 9 = x\)[/tex]
Using the property [tex]\(b^y = z\)[/tex] for logarithms, we have [tex]\((\frac{1}{3})^x = 9\)[/tex].
Recognizing [tex]\(9\)[/tex] as [tex]\(3^2\)[/tex], we rewrite the equation:
[tex]\((3^{-1})^x = 3^2 \)[/tex], which simplifies to [tex]\(3^{-x} = 3^2\)[/tex].
Equating the exponents, [tex]\(-x = 2\)[/tex], so [tex]\(x = -2\)[/tex].
So [tex]\(x = -2\)[/tex].
These steps will lead us to the answers:
- a) [tex]\(32\)[/tex]
- b) [tex]\(2\)[/tex]
- c) [tex]\(1\)[/tex]
- d) [tex]\(4\)[/tex]
- e) [tex]\(2\)[/tex]
- f) [tex]\(-2\)[/tex]
