Question

question 2
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^e = a\\) | \\(\log_b(a) = e\\)
\\(5^2 = 25\\) ✔ | \\(\log_5(25) = 2\\)
\\(8^4 = 4096\\) | \\(\log_8(4096) = 4\\) ✔
\\(3^e = 81\\) | \\(\log_3(81) = 4\\) ×
\\(4^x = 4096\\) × | \\(\log_4(y) = 6\\)
\\(367 = 4^e\\) | \\(\log_4(367) = x\\) ✔
\\(w = k^{12.5}\\) | \\(\log_k(w) = 12.5\\) ✔
\\(bumfuzzle = gardyloo^{bigbanana}\\) | \\(\log_{gardyloo}(bumfuzzle) = bigbanana\\) ✔
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Explanation:

Step1: Analyze \(3^c = 81\)

We know that \(3^4 = 81\) (since \(3\times3\times3\times3 = 81\)). So for the equation \(3^c = 81\), \(c = 4\). Then the logarithmic form \(\log_3(81)=c\), so \(\log_3(81) = 4\) is actually correct? Wait, maybe the system marked it wrong by mistake? Wait, no, let's check the other incorrect one: \(4^x = 4096\).

Step2: Analyze \(4^x = 4096\)

We need to find \(x\) such that \(4^x = 4096\). Let's calculate \(4^5=1024\), \(4^6 = 4096\) (since \(4\times4\times4\times4\times4\times4=4096\)). So \(x = 6\). Then the logarithmic form \(\log_4(4096)=6\), so the given \(\log_4(y)=6\) should have \(y = 4096\), and the exponential form \(4^x = 4096\) has \(x = 6\).

For the \(3^c = 81\) part: Since \(3^4 = 81\), then \(\log_3(81)=4\) is correct. Maybe a system error, but let's focus on the \(4^x = 4096\) and \(3^c = 81\) (if we consider the cross). Wait, the problem is to correct the incorrect parts. Let's handle \(3^c = 81\) first:

For \(3^c = 81\), using the definition \(b^c=a\iff\log_b(a)=c\), so \(3^c = 81\iff\log_3(81)=c\). Since \(3^4 = 81\), \(c = 4\), so \(\log_3(81)=4\) is correct. Now the other incorrect: \(4^x = 4096\). Let's find \(x\):

\(4^1 = 4\), \(4^2 = 16\), \(4^3 = 64\), \(4^4 = 256\), \(4^5 = 1024\), \(4^6 = 4096\). So \(x = 6\). So the exponential form \(4^x = 4096\) has \(x = 6\), and the logarithmic form \(\log_4(4096)=6\). So the incorrect \(4^x = 4096\) (marked wrong) should have \(x = 6\), and the logarithmic form \(\log_4(y)=6\) should have \(y = 4096\).

For the \(3^c = 81\) and \(\log_3(81)=4\): Since \(3^4 = 81\), \(\log_3(81)=4\) is correct. Maybe a mistake in the system's marking. For the \(4^x = 4096\): \(x = 6\), so the exponential form is \(4^6 = 4096\), and the logarithmic form is \(\log_4(4096)=6\).

Answer:

For \(3^c = 81\), \(\log_3(81)=4\) is correct (maybe system error). For \(4^x = 4096\), \(x = 6\) (so \(4^6 = 4096\)) and \(\log_4(4096)=6\).