logarithm problem: getting wrong answer for 2^{2 log_2(x)} - x = 6

[hopefullynottaken]

Can somebody tell me what am i doing wrong here?

\(\displaystyle 2^{2\log_2(x)}\, -\, x\, =\, 6\)

\(\displaystyle log_2\left(2^{2\log_2(x)}\right)\, -\, \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle 2\, \log_2(x)\, \cdot\, \log(2^2)\, -\, \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle 2\, \log_2(x)\, -\, \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle \log_2(x^2):x\, =\, \log_2(6)\)

\(\displaystyle \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle x\, =\, 6\)

 

[mmm4444bot]

Can somebody tell me what am i doing wrong here?

\(\displaystyle 2^{2\log_2(x)}\, -\, x\, =\, 6\)

\(\displaystyle log_2\left(2^{2\log_2(x)}\right)\, -\, \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle 2\, \log_2(x)\, \cdot\, \log(2^2)\, -\, \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle 2\, \log_2(x)\, -\, \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle \log_2(x^2) : x\, =\, \log_2(6)\)

\(\displaystyle \log_2(x)\, =\, \log_2(6)\)

\(\displaystyle x\, =\, 6\)

In your first step, you took the log-base2 of each side, but you did not do it properly, on the left-hand side. We cannot take logarithms term-by-term.

For example, if we take the logarithm of the expression a+b, it is not log(a)+log(b).

It is log(a+b).

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I will suggest a different way to start.

Look at the original exponent: 2*log₂(x)

Use a property of logarithms to rewrite that exponent as: log₂(???).

Try it, and see where that goes. :cool:

 

[tkhunny]

\(\displaystyle 2^{log_{2}(x)} = ??\) Ponder on this and make your life WAY easier.