Solved on Jan 13, 2024

Evaluate the logarithmic expressions: (a) $\log_{5} \frac{1}{25}$ (b) $\log_{2} 32$

STEP 1

Assumptions
1. We are evaluating logarithmic expressions.
2. The base of the logarithm in expression (a) is 5.
3. The argument of the logarithm in expression (a) is $\frac{1}{25}$ .
4. The base of the logarithm in expression (b) is 2.
5. The argument of the logarithm in expression (b) is 32.

STEP 2

To evaluate the logarithm in expression (a), $\log _{5} \frac{1}{25}$ , we need to find the exponent to which the base 5 must be raised to obtain the argument $\frac{1}{25}$ .

STEP 3

Recall the property of logarithms that states $\log_b b^x = x$ . We need to express the argument $\frac{1}{25}$ as a power of 5.

STEP 4

Recognize that $25$ is a power of $5$ , specifically $5^2$ . Therefore, $\frac{1}{25}$ can be written as $5^{-2}$ .

STEP 5

Now we can write the logarithm with the argument as a power of the base:
$\log _{5} \frac{1}{25} = \log _{5} 5^{-2}$

STEP 6

Using the property of logarithms from STEP_3, we find that:
$\log _{5} 5^{-2} = -2$
Thus, $\log _{5} \frac{1}{25} = -2$ .

STEP 7

Next, we evaluate the logarithm in expression (b), $\log _{2} 32$ . We need to find the exponent to which the base 2 must be raised to obtain the argument 32.

STEP 8

Recognize that $32$ is a power of $2$ , specifically $2^5$ .

STEP 9

Now we can write the logarithm with the argument as a power of the base:
$\log _{2} 32 = \log _{2} 2^5$

STEP 10

Using the property of logarithms from STEP_3, we find that:
$\log _{2} 2^5 = 5$
Thus, $\log _{2} 32 = 5$ .
The solutions are: (a) $\log _{5} \frac{1}{25} = -2$ (b) $\log _{2} 32 = 5$

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Evaluate the logarithmic expressions: (a) $\log_{5} \frac{1}{25}$ (b) $\log_{2} 32$

STEP 1

Assumptions
1. We are evaluating logarithmic expressions.
2. The base of the logarithm in expression (a) is 5.
3. The argument of the logarithm in expression (a) is $\frac{1}{25}$ .
4. The base of the logarithm in expression (b) is 2.
5. The argument of the logarithm in expression (b) is 32.

STEP 2

To evaluate the logarithm in expression (a), $\log _{5} \frac{1}{25}$ , we need to find the exponent to which the base 5 must be raised to obtain the argument $\frac{1}{25}$ .

STEP 3

Recall the property of logarithms that states $\log_b b^x = x$ . We need to express the argument $\frac{1}{25}$ as a power of 5.

STEP 4

Recognize that $25$ is a power of $5$ , specifically $5^2$ . Therefore, $\frac{1}{25}$ can be written as $5^{-2}$ .

STEP 5

Now we can write the logarithm with the argument as a power of the base:
$\log _{5} \frac{1}{25} = \log _{5} 5^{-2}$

STEP 6

Using the property of logarithms from STEP_3, we find that:
$\log _{5} 5^{-2} = -2$
Thus, $\log _{5} \frac{1}{25} = -2$ .

STEP 7

Next, we evaluate the logarithm in expression (b), $\log _{2} 32$ . We need to find the exponent to which the base 2 must be raised to obtain the argument 32.

STEP 8

Recognize that $32$ is a power of $2$ , specifically $2^5$ .

STEP 9

Now we can write the logarithm with the argument as a power of the base:
$\log _{2} 32 = \log _{2} 2^5$

STEP 10

Using the property of logarithms from STEP_3, we find that:
$\log _{2} 2^5 = 5$
Thus, $\log _{2} 32 = 5$ .
The solutions are: (a) $\log _{5} \frac{1}{25} = -2$ (b) $\log _{2} 32 = 5$

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Evaluate the logarithmic expressions: (a) log⁡5125\log_{5} \frac{1}{25}log5​251​ (b) log⁡232\log_{2} 32log2​32

STEP 1

Assumptions1. We are evaluating logarithmic expressions.2. The base of the logarithm in expression (a) is 5.3. The argument of the logarithm in expression (a) is 125\frac{1}{25}251​.4. The base of the logarithm in expression (b) is 2.5. The argument of the logarithm in expression (b) is 32.

STEP 2

To evaluate the logarithm in expression (a), log⁡5125\log _{5} \frac{1}{25}log5​251​, we need to find the exponent to which the base 5 must be raised to obtain the argument 125\frac{1}{25}251​.

STEP 3

Recall the property of logarithms that states log⁡bbx=x\log_b b^x = xlogb​bx=x. We need to express the argument 125\frac{1}{25}251​ as a power of 5.

STEP 4

Recognize that 252525 is a power of 555, specifically 525^252. Therefore, 125\frac{1}{25}251​ can be written as 5−25^{-2}5−2.

STEP 5

Now we can write the logarithm with the argument as a power of the base:log⁡5125=log⁡55−2\log _{5} \frac{1}{25} = \log _{5} 5^{-2}log5​251​=log5​5−2

STEP 6

Using the property of logarithms from STEP_3, we find that:log⁡55−2=−2\log _{5} 5^{-2} = -2log5​5−2=−2Thus, log⁡5125=−2\log _{5} \frac{1}{25} = -2log5​251​=−2.

STEP 7

Next, we evaluate the logarithm in expression (b), log⁡232\log _{2} 32log2​32. We need to find the exponent to which the base 2 must be raised to obtain the argument 32.

STEP 8

Recognize that 323232 is a power of 222, specifically 252^525.

STEP 9

Now we can write the logarithm with the argument as a power of the base:log⁡232=log⁡225\log _{2} 32 = \log _{2} 2^5log2​32=log2​25

STEP 10

Using the property of logarithms from STEP_3, we find that:log⁡225=5\log _{2} 2^5 = 5log2​25=5Thus, log⁡232=5\log _{2} 32 = 5log2​32=5.The solutions are: (a) log⁡5125=−2\log _{5} \frac{1}{25} = -2log5​251​=−2 (b) log⁡232=5\log _{2} 32 = 5log2​32=5

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Evaluate the logarithmic expressions: (a) $\log_{5} \frac{1}{25}$ (b) $\log_{2} 32$

Assumptions
1. We are evaluating logarithmic expressions.
2. The base of the logarithm in expression (a) is 5.
3. The argument of the logarithm in expression (a) is $\frac{1}{25}$ .
4. The base of the logarithm in expression (b) is 2.
5. The argument of the logarithm in expression (b) is 32.

To evaluate the logarithm in expression (a), $\log _{5} \frac{1}{25}$ , we need to find the exponent to which the base 5 must be raised to obtain the argument $\frac{1}{25}$ .

Recall the property of logarithms that states $\log_b b^x = x$ . We need to express the argument $\frac{1}{25}$ as a power of 5.

Recognize that $25$ is a power of $5$ , specifically $5^2$ . Therefore, $\frac{1}{25}$ can be written as $5^{-2}$ .

Now we can write the logarithm with the argument as a power of the base:
$\log _{5} \frac{1}{25} = \log _{5} 5^{-2}$

Using the property of logarithms from STEP_3, we find that:
$\log _{5} 5^{-2} = -2$
Thus, $\log _{5} \frac{1}{25} = -2$ .

Next, we evaluate the logarithm in expression (b), $\log _{2} 32$ . We need to find the exponent to which the base 2 must be raised to obtain the argument 32.

Recognize that $32$ is a power of $2$ , specifically $2^5$ .

Now we can write the logarithm with the argument as a power of the base:
$\log _{2} 32 = \log _{2} 2^5$

Using the property of logarithms from STEP_3, we find that:
$\log _{2} 2^5 = 5$
Thus, $\log _{2} 32 = 5$ .
The solutions are: (a) $\log _{5} \frac{1}{25} = -2$ (b) $\log _{2} 32 = 5$