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Math problem:

Find $x$, given that $ \, 2^2 \times 2^4 \times 2^6 \times 2^8 \times \ldots \times 2^{2x} = \left( 0.25 \right)^{-36}$

To solve this question, I changed the left side of the equation to $2^{2+4+6+ \ldots + 2x}$ and the right side to: $\frac{2^{74}}{3^{36}}$.

My question is how can $3$ to a power (in this case $36$) be changed to $2$ to a power? (algebraically-without a calculator)

By checking with a calculator and doing $\log$, I found that it is not a whole number and therefore the wrong method for this question.

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    $\begingroup$ $0.25 = 1/4$ which is a power of $2$. $\endgroup$ – wgrenard Oct 12 '17 at 21:20
  • $\begingroup$ You can certainly write $3^{36}=2^{\frac{36\ln 3}{\ln 2}}$, but your substitution of the right hand side is unwarranted and ultimately wrong. $\endgroup$ – user228113 Oct 12 '17 at 21:20
  • $\begingroup$ $2^{\log_3 n}$? (Which happens to be equivalent to $n^{\log_3 2} = n^{\log 2 / \log 3}$.) $\endgroup$ – Daniel Schepler Oct 12 '17 at 21:22
  • $\begingroup$ you only have power of two on the right side and its $2^{72}$ $\endgroup$ – Isham Oct 12 '17 at 21:25
  • $\begingroup$ Please take a look at this post to learn about MathJax formatting: math.meta.stackexchange.com/questions/5020/… Using MathJax makes your question easier to read and become definitely more aesthetically pleasing :) (I edited the question for you this time) $\endgroup$ – bertalanp99 Oct 12 '17 at 21:38
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use that $$2+4+6+8+...+2x=2^{x(x+1)}$$ and $$\left(\frac{1}{4}\right)^{-36}=2^{72}$$ and you will have $$2^{2^{x(x+1)}}=2^{72}$$

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  • $\begingroup$ left side of first equation should be in an exponent, or right side should be simply $x(x+1)$. $\endgroup$ – Lubin Oct 12 '17 at 21:25
  • $\begingroup$ i have written down only the exponents $\endgroup$ – Dr. Sonnhard Graubner Oct 12 '17 at 21:28
  • $\begingroup$ I don’t care: as it stands now, your first equality is false. $\endgroup$ – Lubin Oct 13 '17 at 4:58
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$( 0.25)^{-36}=( \frac 1 4) ^{-36}=(2^{-2})^{-36}=(2^{36})^2$

$2^22^42^6....2^{2x}=(2^12^22^32^4...2^x)^2$

So we must have:

$(2^{36})^2=(2^12^22^32^4...2^x)^2$

Or simply:

$2^{36}=2^12^22^32^4...2^x$

$1+2+3+ ....x=36$

$\frac {(x+1)x} 2=36$

$x^2+x=72$

$x=8$

Check $8^2+8=72$

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  • $\begingroup$ -9 is also a solution of the quadratic equation but since the exponant is positif so only 8 is a solution $\endgroup$ – Isham Oct 12 '17 at 21:55
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Take the base-2 logarithm of both members, and you get

$$2+4+6+\cdots+2x=(-2)(-36)$$

or

$$1+2+3+\cdots+x=36.$$

$36$ is the eighth triangular number.


Even though this is irrelevant to the given problem, you convert a power of $2$ to a power of $3$ by writing

$$2^a=3^b,$$ and taking the logarithm (in any base),

$$a\log2=b\log 3$$ or

$$b=a\frac{\log 2}{\log 3}.$$

For integer $a$, $b$ can never be an integer (nor a rational).

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