Question

score on last try: 2.86 of 4 pts. see details for more. at least one scored part is incorrect. jump to first changable incorrect part. > next question get a similar question you can retry this question below practice changing between exponential form and logarithmic form using the table below: note: \\(\log_b (a)\\) can be written using an underscore like this: \\(\log\\_b(a)\\). \\(b^c = a\\) \\(\log\\_b(a) = c\\) \\(5^2 = 25\\) \\(\log_5 (25) = 2\\) \\(8^4 = 4096\\) \\(\log_8 (4096) = 4\\) \\(3^c = 81\\) \\(\log_3 (81) = 4\\) \\(4^6 = 4096\\) \\(\log_4 (y) = 6\\) \\(367 = 4^x\\) \\(\log_4 (367) = x\\) \\(w = k^{12.5}\\) \\(\log_k (w) = 12.5\\) \\(bunfuzzle = gardyloo^{bigbanana}\\) \\(\log\\_gardyloo (bunfuzzle) = bigbanana\\)

Explanation:

Step1: Analyze \( \log_3(81) \)

Recall the definition of a logarithm: \( \log_b(a) = c \) is equivalent to \( b^c = a \). For \( \log_3(81) \), we need to find \( c \) such that \( 3^c = 81 \).
We know that \( 3^4 = 81 \) (since \( 3\times3\times3\times3 = 81 \)). So \( \log_3(81) = 4 \), which means the original marking of \( \log_3(81)=4 \) as incorrect was wrong; it should be correct.

Step2: Analyze \( 4^6 \)

Calculate \( 4^6 \): \( 4^1 = 4 \), \( 4^2 = 16 \), \( 4^3 = 64 \), \( 4^4 = 256 \), \( 4^5 = 1024 \), \( 4^6 = 4096 \). So \( 4^6 = 4096 \) is correct, and the original marking of it as incorrect was wrong.

Step3: Analyze \( \log_4(y) = 6 \)

Using the logarithmic - exponential equivalence \( \log_b(a)=c\Leftrightarrow b^c = a \), for \( \log_4(y)=6 \), we have \( y = 4^6 \). Since \( 4^6 = 4096 \), then \( y = 4096 \), so \( \log_4(4096)=6 \) is correct.

Step4: Analyze \( 3^c = 81 \)

As we found earlier, \( 3^4 = 81 \), so \( c = 4 \), and \( 3^c = 81 \) (with \( c = 4 \)) is correct.

Step5: Analyze \( 8^4 \)

Calculate \( 8^4 \): \( 8^2=64 \), \( 8^4=(8^2)^2 = 64^2=4096 \), so \( 8^4 = 4096 \) is correct.

Step6: Analyze \( \log_8(4096) \)

Let \( \log_8(4096)=x \), then by the logarithmic - exponential equivalence, \( 8^x = 4096 \). We know that \( 8^4=(2^3)^4 = 2^{12}=4096 \), so \( x = 4 \), and \( \log_8(4096)=4 \) is correct.

Step7: Analyze \( 5^2 \)

\( 5^2=5\times5 = 25 \), so \( 5^2 = 25 \) is correct, and \( \log_5(25) \): let \( \log_5(25)=x \), then \( 5^x = 25 \), and \( x = 2 \), so \( \log_5(25)=2 \) is correct.

Step8: Analyze \( 367 = 4^x \)

We know that \( 4^4 = 256 \), \( 4^5 = 1024 \). Since \( 256<367<1024 \), there is no integer \( x \) such that \( 4^x = 367 \). But if we consider real - valued logarithms, \( x=\log_4(367)=\frac{\ln(367)}{\ln(4)}\approx\frac{5.99}{1.39}\approx4.31 \), but the problem seems to be about integer exponents (from the context of other problems with integer exponents). So \( 367 = 4^x \) has no integer solution, and \( \log_4(367)=x \) (if we consider real \( x \)) is a non - integer, but from the marking, if we assume integer exponents, it's incorrect.

Step9: Analyze \( w = k^{12.5} \) and \( \log_k(w)=12.5 \)

By the logarithmic - exponential equivalence \( \log_b(a)=c\Leftrightarrow b^c = a \), if \( \log_k(w)=12.5 \), then \( w = k^{12.5} \), so this pair is correct.

Step10: Analyze \( bumfuzzle = gardyloo^{bigbanana} \) and \( \log_{gardyloo}(bumfuzzle)=bigbanana \)

By the definition of a logarithm, \( \log_b(a)=c\Leftrightarrow b^c = a \). Here, if we let \( b = gardyloo \), \( a = bumfuzzle \), and \( c = bigbanana \), then \( \log_{gardyloo}(bumfuzzle)=bigbanana \) is equivalent to \( gardyloo^{bigbanana}=bumfuzzle \), so this pair is correct.

Answer:

The incorrect markings were for \( \log_3(81) = 4 \), \( 4^6 = 4096 \), and the correct relationships are as derived above (e.g., \( \log_3(81)=4 \) is correct, \( 4^6 = 4096 \) is correct, etc.). For the equation \( 367 = 4^x \), there is no integer \( x \) such that \( 4^x=367 \).