Jika 4^6x=8000 maka (4^x+1)^2= __

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Jika 4^6x=8000 maka (4^x+1)^2= __

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Answer 1

The process of working out this problem would therefore consist of first converting the question to logarithmic and then use logarithmic principles to find the answer, once that step is done, substituting it into secondary equation, use binomial expansions to finally derive at the result. By working through the process one can find that s (expressed in logarithms), amounts in s = 2^((2/3)*log4(8000)) + 2* 1* 2^((1/3)*log_4(8000)) + 1.

Answer 2

Step 1: From the given equation, 4^6x=8000, we find that 6x is the log of 8000 to base 4. We denote y = 2x.
Step 2: Therefore, we rewrite the given equation to be 4^y = 8000.
Step 3: Next, we use the relationship between logarithm and exponential to convert this equation into logarithm and behooves us to take the base preserve the equation as log 8000, and the base as 4. This gives us y = log 8000 to base 4.
Step 4: Now substitute y = 2x in the second equation (4^x+1)^2 to get (2^(y/3)+1)^2.
Step 5: Use binomial expansion to solve the square on the right. We multiply every term in the binomial (a+b)^2 by itself and add together, according to the rule (a+b)^2 = a^2 + 2ab + b^2, obtaining the final value of s, (2^((2/3)*y) + 2* 1*2^(y/3) + 1).

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Jika 4^6x=8000 maka (4^x+1)^2= __

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