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This is the exercise, there are no clues in the book about it. $$ 40\log_{4x}x^\frac{1}{2}-14\log_{16x}x^3=-\log_{\frac{1}{2}x}x^2 $$

Solutions given by the book: $x=1; x=4; x=\frac{\sqrt{2}}{2}.$

And this is what I did so far:

  1. conditions for existence: $$ \Biggl\{ \begin{eqnarray} x &\gt& 0.\\ x &\ne& \frac 14; x \ne \frac{1}{16}; x \ne 2. \end{eqnarray} $$
  2. I simplified some exponent in the arguments: $$ 20\log_{4x}x-42\log_{16x}x+2\log_{\frac{1}{2}x}x=0 $$
  3. Change of bases: $$ \frac{20}{\log_x4x}-\frac{42}{\log_x16x}+\frac{2}{\log_x\frac{1}{2}x}=0 $$
  4. Observing that: $\log_xnx=1+\log_xn$
  5. least common multiple: $$ \frac{20(1+\log_x16)(1+\log_x\frac12)-42(1+\log_x4)(1+\log_x\frac12)+2(1+\log_x4)(1+\log_x16)}{(1+\log_x16)(1+\log_x4)(1+\log_x\frac12)}=0 $$
  6. denominator can be toggled, as for conditions for existence
  7. then I lost confidence in what I was doing...

Any clue is welcome, thanks :]

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  • $\begingroup$ Try substituting $u = \log_2{x}$. $\endgroup$ Oct 7 '19 at 19:11
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We can use that

$$\log_a^b=\frac{\log a}{\log b}$$

therefore

$$40\log_{4x}x^\frac{1}{2}-14\log_{16x}x^3=-\log_{\frac{1}{2}x}x^2$$

$$40\frac{\log x^\frac{1}{2}}{\log {4x}}-14\frac{\log x^3}{\log {16x}}=-\frac{\log x^2}{\log {\frac{1}{2}x}}$$

$$20\frac{\log x}{\log {x}+\log 4}-42\frac{\log x}{\log {x}+\log 16}=-2\frac{\log x}{\log {x}+\log \frac12}$$

and eliminating $\log x \neq 0$ (which is a solution)

$$\frac{10}{\log {x}+2\log 2}-\frac{21}{\log {x}+4\log 2}=-\frac{1}{\log {x}-\log 2}$$

then let $y=\log x$ and $a=\log 2$ to obtain

$$\frac{10}{y+2a}-\frac{21}{y+4a}+\frac{1}{y-a}=0$$

$$\frac{10(y+4a)(y-a)-21(y+2a)(y-a)+(y+2a)(y+4a)}{(y+2a)(y+4a)(y-a)}=0$$

$$\frac{10(y^2+3ay-4a^2)-21(y^2+ay-2a^2)+(y^2+6ay+8a^2)}{(y+2a)(y+4a)(y-a)}=0$$

$$\frac{-10 y^2+15 ay+10a^2}{(y+2a)(y+4a)(y-a)}=0$$

$$\frac{-5(y-2a)(2y+a)}{(y+2a)(y+4a)(y-a)}=0$$

that is

  • $\log x = 2\log 2$

  • $2\log x = -\log 2$

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  • $\begingroup$ Yes, this is the procedure I'll use for future reference. I'm rewriting it in my notebook, but I'll directly choose to use $\log_2$ in the substitution, because this will simplify a. Thank you for your time. $\endgroup$
    – sabi
    Oct 8 '19 at 9:27
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Nice, I wanted to try myself. I worked the first step as you did, getting to

$$\frac{20}{\log_x4x}-\frac{42}{\log_x16x}+\frac{2}{\log_x\frac{1}{2}x}=0,$$

valid if $x\neq 1$, which is anyhow a solution to the original equation.

Then, dividing by $2$ and rewriting the logarithms as you did, yields

$$\frac{10}{1+\log_x4}-\frac{21}{1+\log_x 16}+\frac1{1-\log_x 2}=0.$$

From here I proceded inverting base and arguments and replacing $\log_{16} x$ with $t$, for simplicity

$$\frac{10}{1+\frac1{2t}}-\frac{21}{1+\frac1{t}}+\frac1{1-\frac1{4t}}=0,$$

which, again for $t\neq 0$, is equivalent to

$$\frac{20}{2t+1}-\frac{21}{t+1}+\frac4{4t-1}=0.$$

Least common denominator brings you to the quadratic equation

$$16t^2-6t-1=0,$$

with the desidered solutions, i.e. $t=\frac12$ and $t=-\frac18$.

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  • $\begingroup$ Excellent! That’s the path I was trying hard. Can you please tell where can I find more info on the substitution? It’s not so clear for me that if $\log_{16}x=t$ then $\log_{4}x=2t$. Would it be the same if I had $\log_{16x}x=t$ and $\log_{4x}x=?$ $\endgroup$
    – sabi
    Oct 8 '19 at 4:43
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    $\begingroup$ @sabi if you want to familiarize with logarithm properties, you should recall that they are just exponents. So, e.g., if $\log_{16}x = \alpha$ and $\log_4x = \beta$, then $16^\alpha=4^\beta =x$, and from there you derive your identity. I can't think of any book in english that deals with this material, right now. Can you give me reference to the book you got this exercise from? $\endgroup$
    – dfnu
    Oct 8 '19 at 8:58
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    $\begingroup$ pearson.it/opera/pearson/0-6117-analisi_matematica_i an Italian book. $\endgroup$
    – sabi
    Oct 8 '19 at 9:29
  • $\begingroup$ @sabi don't worry, I'm Italian, too. If you like, have a look at the exercises I propose www.dfnu.xyz. $\endgroup$
    – dfnu
    Oct 8 '19 at 10:00
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    $\begingroup$ thanks a lot, I'm already checking your website :] $\endgroup$
    – sabi
    Oct 8 '19 at 10:08

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