# Solve the equation $\log_2(\frac{8}{2^x}-1)=x-2$

This equation comes from the book of Prilepko. It looks sufficiently simple but I want to know if my solution is correct.

$\log_2(\frac{8}{2^x}-1)=x-2$

I have only been able to transform into this:

$\log_2(\frac{8-2^x}{2^x})=x-2$

$\iff$ $\log_2(8-2^x)=2x-2$

$\iff$ $8-2^x=2^{2x-2}$

$\iff$ $32-4.2^x-2^{2x}= 0$

Let $t=2^x$

$-t^2-4t+32=0$

This equation will produce $t=-8$ and $t=4$

Therefore $x=2$.

Another question that I wish to ask is when solving these logarithmic equations, one very often substitute a dummy variable to overcome the restriction caused by the order of operations. For example, you cannot do much directly with $5^{2x}-130.5^x+625$. But let $t=5^x$, then you can easily factorize the this equation into $(t-125)(t-5)=0$, hence $(5^x-125)(5^x-5)=0$. Isn't this a sort of manipulating symbols according to certain prescribed rules? Is there any logical difficulty with this kind of operation? Since by substituting a new variable, it means that this variable must inherit all properties which the object of the substitution originally possesses. For example, if $a^x$ is the object of substitution for $t$, then if $a<0$ and x can assumed a form of rational power, doesn't this make no sense at all?

Do you know any example in elementary mathematics where this method of substitution can yield contradictory result?

• Your solution looks correct for the first part, and I believe that is the nicest solution. – Carl Schildkraut Mar 15 '18 at 19:48
• there is just a typo $$\log_2\left(\frac{8-2^x}{\color{red}{2^x}}\right)=x-2$$ – gimusi Mar 15 '18 at 19:48
• Thanks, I have corrected it. But is this method of substitution justified logically? Is there circumstances where you apply substitution and it yields a contradictory result? – James Warthington Mar 15 '18 at 19:51
• Sometimes we get extraneous solutions... solutions to an equation we got from the original equation but sometimes their domains differ. – randomgirl Mar 15 '18 at 19:56
• If $x$ is a real number then $2^x$ is a positive real number. Making the substitution that $t = 2^x$ then $t$ must be a positive real. And if the algebra showed a solution that worked where t was a negative real, it is still does not a solution for $x.$ And if your algebra lead you to the conclusion that solutions only existed for negative values of $t.$ Then there is no solution for real values of $x.$ What is all of this talk of "real numbers" there are numbers that are not real... complex numbers.. for which $2^z$ may be negative. – Doug M Mar 15 '18 at 19:58

## 6 Answers

Your solution is fine, well done. Note that you can omit the second step by doing $$\log_2\left(\frac{8}{2^x}-1\right)=x-2\implies8\cdot2^{-x}-1=2^{x-2}\implies8-2^x=2^{2x-2}$$

A substitution of the form $t=a^x$ for (positive) $a$ can be used at any time without doubting whether it would yield unwanted solutions, since if $t$ is negative then it can be automatically rejected.

• How does this answer the question? – user535339 Mar 15 '18 at 19:52
• Well thanks for checking my answer. But do you know any kind of problems which by substituting a dummy variable, you may get a result that doesn't make sense? – James Warthington Mar 15 '18 at 19:53
• @idk The OP asks, "I want to know if my solution is correct." TheSimpliFire answers, "Your solution is fine." – saulspatz Mar 15 '18 at 19:55
• @saulspatz That was my mistake, I didn't read the first part of the question. – user535339 Mar 15 '18 at 19:57
• Thank you very much! I have no other questions. – James Warthington Mar 15 '18 at 20:07

First of all, yes your solution is correct and you can also check it by putting it again in given equation. And secondly, about your second question, there is nothing wrong in replacing a value by a variable and no ambiguity will be created in this process. You can replace it with even more or less complex variable and it will remain to its own properties.

Your answer $x=2$ is correct. You have made minor typo's which you can fix.

I let y=2^x and substituted in $$2^{2x-2}=8-2x$$ to get $$y^2 +4y-32=0$$

Solving for $y$ and we get $y=4$ or $y=-8$

Clearly, $y=-8$ is not acceptable.

Thu $y=4$ which implies $x=2$

If you want "to stretch" the logarithm method or solution to the very end here it is:

$$\log_{2} \left ( \frac{8}{2^{x}} - 1 \right ) = x-2$$

$$\log_{2} \left ( \frac{8-2^{x}}{2^{x}} \right ) = x-2$$

$$\log_{2} \left (8-2^{x} \right ) - \log_{2} \left ( 2^{x} \right ) = x-2$$

$$\log_{2} \left (2^{3}-2^{x} \right ) - x = x-2$$

$$\log_{2} \left (2^{3}-2^{x} \right ) = 2x-2$$

$$\log_{2} \left (2^{3} \left ( 1- 2^{x-3}\right ) \right ) = 2x-2$$

$$\log_{2} \left (2^{3}\right ) + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2$$

$$3\log_{2} \left (2\right ) + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2$$

$$3 + \log_{2} \left ( 1- 2^{x-3} \right ) = 2x-2$$

$$\log_{2} \left ( 1- 2^{x-3} \right ) = 2x-5$$

Using the $\textrm{Antilogarithm}$ (see here or here for any details)

$$\textrm{antilog}_{2} \left ( \log_{2} \left ( 1- 2^{x-3} \right ) \right ) = \textrm{antilog}_{2} \left ( 2x-5 \right )$$

$$1- 2^{x-3}= 2^{2x-5}$$

Rearranging the above equation translates into:

$$1- 2^{x}\times 2^{-3}= 2^{2x}\times 2^{-5}$$

$$2^{2x}\times 2^{-5} - 2^{x}\times 2^{-3} - 1 = 0$$

Multiplying by $2^{5}$ becomes into the expression you had found:

$$2^{2x} - 2^{x}\times 2^{2} - 32 = 0$$

From then on it is consistent with a quadratic equation:

$$A=2^{x}$$

$$A_{1,2}=\frac{-4\pm \sqrt{16+128}}{2}=\frac{-4\pm \sqrt{144}}{2}$$

$$A_{1,2}=\frac{-4\pm 12}{2}\,,A_{1}=\frac{-4+12}{2}=4\,,A_{2}=\frac{-4-12}{2}=-8$$

Since we cannot obtain negative results from a power the second solution is discarded. Therefore:

$$2^{x}=4$$

$$x=2$$

Becoming the result you had just found. So it is correct. As mentioned this is more or less the same of what others before me had just commented.

observe that $8=2^3$.

$$\ln (2^{3-x}-1)=(x-2)\ln (2)=\ln (2^{x-2})$$

$$2^{3-x}-2^{x-2}=1$$

Put $2^x=t >0$.

it becomes $$8/t-t/4-1=0$$

$$t^2+4t-32=0$$ $t=4 \implies x=2$

It is correct and the best method to solve this kind of equations, fix just a typo

$$\log_2\left(\frac{8-2^x}{\color{red}{2^x}}\right)=x-2$$

Regarding to the general question note that the substitution $a^x=t$ is nice since we are not limiting the possible values for $a^x>0$. Of course we can obtain some not positive solution but this is not a problem since we can simply discard them as you have done in this example. Otherwise it should be uncorrect substitute a variable $t$ which can assume also negative values with $a^x$ because in this way we could loss solutions.

• How does this answer the question? – user535339 Mar 15 '18 at 19:46
• ah ok I didn't recognize that it was just a typo, sorry! – gimusi Mar 15 '18 at 19:47