# Problem with logarithms

Problem:

Solve:

$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$

My attempt:

$$\frac{1}{2^x} = \frac{5}{8^{x+2}}$$

$$\Rightarrow 5 \cdot 2^x = 8^{x+2}$$

$$\Rightarrow 2^{\log_2 5+x} = 8^{x+2}$$

$$\Rightarrow (\log_2 5 + x)(\log_a 2) = (x+2)(\log_a 8)$$

And then just keep going like this, but I'm obviously wrong already as the answer is:

$$x = \frac{\ln 5 - 9 \ln 2}{2 \ln 2}$$

What am I doing wrong?

EDIT: If someone could enlarge the latex for me that would be great.

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you have a good start, but you can continue better. See my solution. –  Babak Miraftab Feb 21 '14 at 20:13
You can get the larger displayed latex form by bracketing the math code between two pairs of dollar signs (insead of one pair of dollar signs). I just did this for your question. –  Dave L. Renfro Feb 21 '14 at 20:17
Thank you for that! –  user3200098 Feb 21 '14 at 20:21
With the content of the answers below, I'm not certain that the given answer is correct. Is there perhaps a copying error? –  abiessu Feb 21 '14 at 21:41

$2^{\log_2 5+x} = 8^{x+2}$ implies that $2^{\log_2 5+x} = 2^{3x+6}$ and so $\log_2 5+x=3x+6$. We can conclude that $x=\frac{\log_2 5-6}{2}$.

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Thank you...!!! –  user3200098 Feb 21 '14 at 20:20

Some of the answers don't mention that $8=2^3$. $$5\cdot 2^x = 8^{x+2}$$

Remember that $8=2^3$. So $$5\cdot 2^x = (2^3)^{x+2}.$$ $$5\cdot 2^x = 2^{3(x+2)}$$ $$5\cdot 2^x = 2^{3x+6}$$ $$5 = 2^{3x+6}\cdot 2^{-x} = 2^{2x+6}$$ $$\log_2 5 = 2x+6$$ $$-6 + \log_2 5 = 2x$$ $$\frac{-6+\log_2 5}{2} = x$$

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Thanks a lot :) –  user3200098 Feb 21 '14 at 22:47
I didn't realize I could do $5 = 2^{3x+6}\cdot 2^{-x} = 2^{2x+6}$ though. Multiply by negative exponent on both sides that is to get rid of the $2^x$ on one side. –  user3200098 Feb 21 '14 at 22:52
@user3200098 : What is being done is this: $2^{3x+6}\cdot 2^{-x} = 2^{\Big(3x+6\Big)+\Big(-x\Big)}$. Then simplify $(3x+6)+(-x)$ to get $2x+6$. –  Michael Hardy Feb 22 '14 at 0:34

With a slight modification, we have

$$5\cdot 2^x=8^{x+2}=(2^3)^{x+2}=2^{3x+6}$$

which means that

$$5=2^{2x+6}\\ \log_2 5=2x+6\\ x=\frac{\log_2 (5)-6}2$$

Transforming to the natural logarithm would look like

$$5=2^{2x+6}\\ \ln 5=(2x+6)\ln 2\\ x=\frac{\ln 5-6\ln 2}{2\ln 2}$$

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HINT Use substitution to solve the equation easily $\to$ pose $2^x=t$

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