A logarithmic equation

We write $\log_x16$ as $4\log_x2$ , and $\log_{16}x$ as $\frac1{\log_x16} = \frac1{4\log_x2}$ . Similarly, $\log_x{512} = 9\log_x2$ , and $\log_{512}x = \frac1{9\log_x2}$ . Hence, $\begin{aligned} 4\log_x2 + \frac1{4\log_x2} &=& 9\log_x2 + \frac1{9\log_x2} \\ \left(\frac14 - \frac19\right)\left(\frac1{\log_x2}\right) &=& 5\log_x2 \\ \frac1{36} &=& (\log_x2)^2 \\ \pm\frac16 &=& \log_x2. \end{aligned}$ Hence, $x = 64$ or $x = \frac1{64}$ , and the sum of these is $\frac{4097}{64}$ . We obtain $4097+64 = 4161$ , and the last three digits are $\boxed{161}$ .

Let $y = \log_2 x.$ Then $\dfrac{y}{4} = \dfrac{\log_2 x}{4} = \dfrac{\log_2 x}{\log_2 16} = \log_{16} x$ and $\dfrac{y}{9} = \dfrac{\log_ 2 x }{9} = \dfrac{\log_2 x}{\log_2 512} = \log_{512} x.$ In addition, $\log_x 16 = \dfrac{1}{\log_{16} x} = \dfrac{4}{y}$ and $\log_x 512 = \dfrac{1}{\log_{512} x} = \dfrac{9}{y}.$ Therefore, the given equation becomes $\dfrac{y}{4} + \dfrac{4}{y} = \dfrac{y}{9} + \dfrac{9}{y}.$ Combining fractions on both sides gives $\dfrac{y^2+16}{4y} = \dfrac{y^2+81}{9y},$ and cross-multiplying, we have $9y(y^2+16) = 4y(y^2+81).$ If $y = 0,$ then $x = 2^0 = 1,$ which is not a valid solution since $\log_x 16$ is not defined if $x = 1.$ Thus, we can divide both sides by $y$ to obtain the following: $\begin{aligned} 9(y^2+16) &= 4(y^2+81) \\ 9y^2+144 &= 4y^2+324 \\ 5y^2 &= 180 \\ y^2 &= 36 \\ y &= \pm 6. \end{aligned}$ Therefore, either $\log_2 x = 6,$ so $x = 2^6 = 64,$ or $\log_2 x = -6,$ so $x = 2^{-6} = \dfrac{1}{64}.$ The sum of these two solutions is $64 + \dfrac{1}{64} = \dfrac{4097}{64},$ and the last three digits of $4097+64 = 4161$ are $\boxed{161}.$

Using the identity that -

$log_b a = \frac{ log_m a}{log_m b}$ ,

Taking the bases as 2 [As we see that $16$ and $512$ are powers of $2$ ] ,

and thus our simplified equation will be -

$\frac{log_2 x}{log_2 16} + \frac{log_2 16}{log_2 x} = \frac{log_2 x}{log_2 512} + \frac{log_2 512}{log_2 x}$

$\Rightarrow \frac{log_2 x}{4} + \frac{4}{log_2 x} = \frac{log_2 x}{9} + \frac{9}{log_2 x}$

$\Rightarrow \frac{log_2 x}{4} - \frac{log_2 x}{9} = \frac{9}{log_2 x} - \frac{4}{log_2 x}$

$\Rightarrow (log_2 x )(\frac{1}{4} - \frac{1}{9}) = (\frac{1}{log_2 x}) (9-4)$

$\Rightarrow (log_2 x)(\frac{5}{36}) =(\frac{5}{log_2 x})$

$\Rightarrow (log_2 x)(\frac{1}{36}) =(\frac{1}{log_2 x})$

Now by transposing the constant terms to one side ,

$\Rightarrow (log_2 x)^2 = 36$

$\Rightarrow log_2 x = \pm 6$

$\Rightarrow x = 2^{\pm 6}$

Thus , we get $x = 2^6 = 64$ and $x = 2^{-6} = \frac{1}{64}$

Thus the sum of values of $x$

$= 64 + \frac{1}{64} = \frac{4097}{64}$

Thus $a + b = 4097 + 64 = 4161$

And hence , last three digits are $161$

Cheers!

We let $log_{16} x=k$ . Then $log_x 16=\frac{1}{k}$ . Also, since $log_{512} 16=log_{2^9} 2^4=\frac{4}{9}$ , $log_{512} x=(log_{16} x)(log_{512} 16)=(k)(\frac{4}{9})=(\frac{4k}{9})$ . Hence $log_x 512=\frac{9}{4k}$ .

Now we solve the equation. In terms of $k$ , the equation becomes $k+\frac{1}{k}=\frac{4k}{9}+\frac{9}{4k}$ . Thus $k-\frac{4k}{9}=\frac{9}{4k}-\frac{1}{k}$ . Hence $\frac{5k}{9}=\frac{5k}{4k^2}$ . Thus $4k^2=9$ , and $k^2=\frac{9}{4}$ . $k=\pm \frac{3}{2}$ .

If $k=\frac{3}{2}$ , then $x=16^{\frac{3}{2}}=64$ . If $k=-\frac{3}{2}$ , $x=\frac{1}{16^{\frac{3}{2}}}=\frac{1}{64}$ .

Thus the sum of all positive solutions is $\frac{1}{64}+64=\frac{64^2+1}{64}=\frac{4097}{64}$ .

Hence $a=4097$ and $b=64$ . $a+b=4161$ , so the last $3$ digits of $a+b$ is $\boxed{161}$ .

$\begin{aligned} \log_{16} x + \log_x 16 & = \log_{512} x + \log_x 512 \\ \frac{\log_2 x}{\log_2 16} + \frac{\log_2 16}{\log_2 x} & = \frac{\log_2 x}{\log_2 512} + \frac{\log_2 512}{\log_2 x} \\ \frac{\log_2 x}{4} + \frac{4}{\log_2 x} & = \frac{\log_2 x}{9} + \frac{9}{\log_2 x} \\ \left(\frac{1}{4} - \frac{1}{9} \right) \log_2 x & = \frac{9-4}{\log_2 x} \\ \left(\frac{5}{36} \right) \log_2 x & = \frac{5}{\log_2 x} \\ \left(\log_2 x\right)^2 & = 36 \\ \Rightarrow \log_2 x & = \pm 6 \\ x & = \begin{cases} 2^6 = 64 \\ 2^{-6} = \frac{1}{64} \end{cases} \\ 64 + \frac{1}{64} & = \frac{4097}{64} \end{aligned}$

$\Rightarrow a + b = 4097+64 = 4\boxed{161}$

We may rewrite the equation this way:

$\log_{16} x -\log_{512} x = \log_x 512 - \log_x 16$

Note that the RHS is equal to $\log_x \frac{512}{16} = \log_x 32$

We can change the base of every expression on the LHS to $16$

We have the LHS equal to $\log_{16} x -\frac{\log_{16} x}{\log_{16} 512}$

Note that $\log_{16} 512 = \frac{9}{4}$ . We have the equation:

$\log_{16} x - \frac{4}{9}\log_{16} x = \log_x 32$ or $\frac{5}{9}\log_{16} x = \log_x 32$

We may change the base of the RHS to $16$ and thus:

$\log_x 32 = \frac{\log_{16} 32}{\log_{16} x}$ . We now have the equation equal to:

$\frac{5}{9}(\log_{16} x)^{2} = \log_{16} 32$ , but $\log_{16} 32 =\frac{5}{4}$

Therefore: $(\log_{16} x)^{2} = \frac{9}{4} \Rightarrow \log_{16} x = \pm\frac{3}{2}$

$x = 4^{\pm 3}$ so the sum of all solutions is $64 + \frac{1}{64} = \frac{4097}{64}$

$a + b = 4161$ , $161$ is the answer.

$log_{16}x + log_{x}16 = log_{512}x + log_{x}512$

$\frac{log_{2}x}{log_{2}2^{4}} + \frac{log_{2}2^{4}}{log_{2}x} = \frac{log_{2}x}{log_{2}2^{9}} + \frac{log_{2}2^{9}}{log_{2}x}$

$\frac{log_{2}x}{4} + \frac{4}{log_{2}x} = \frac{log_{2}x}{9} + \frac{9}{log_{2}x}$

$\frac{log_{2}x}{4} - \frac{log_{2}x}{9} = \frac{9}{log_{2}x} - \frac{4}{log_{2}x}$

$\frac{5 \times log_{2}x}{36} = \frac{5}{log_{2}x}$

$(log_{2}x)^{2} = 36$

$log_{2}x = 6$ or $log_{2}x = -6$

$x = 2^{6}$ or $x = 2^{-6}$

$64 + \frac{1}{64} = \frac{4096 + 1}{64} = \frac{4097}{64}$

So $a = 4097$ and $b = 64$

$a+b = 4161$

And the last three digits are $161$

$\log_{16} x + \log_x 16 = \log_{512} x + \log_x 512.$ Using the change-of-base rule, we have $\dfrac {\log_2 x}{\log_2 16} + \dfrac {\log_2 16}{\log_2 x} = \dfrac {\log_2 x}{\log_2 512} + \dfrac {\log_2 512}{\log_2 x},$ so $\dfrac {\log_2 x}{4} + \dfrac {4}{\log_2 x} = \dfrac {\log_2 x}{9} + \dfrac {9}{\log_2 x}.$ Simplifying, we get $\dfrac {5}{36} \log_2 x = \dfrac {5}{\log_2 x} \implies \left( \log_2 x \right)^2 = 36$ .

$\implies \log_2 x = \pm 6 \implies x = 2^{\pm 6}$ .

$\implies x \in \left\{ 64, \dfrac {1}{64} \right\}$ .

$\implies \displaystyle\sum x = 64 + \dfrac {1}{64} = \dfrac {4097}{64}$ , so our answer is $(4097+64) \bmod 1000 \equiv 4161 \bmod 100 = \boxed {161}$ .

Let $\log_2 x = t$ then we can write the equation as: $t/4 + 4/t = t/9 + 9/t$ $t^2 = 36$ So we have $t = 6$ or $t = -6$ which means $x = 64$ or $x = 1/64$

There are certain laws of logarithms required for this question. I shall list them below.

$\log _ab^{n} = n\log _ab$

$\log_aa = 1$

$\log _ab = \frac{\log _cb}{\log _ca}$ where c can be any number whatsoever. In this case we will chose c = 2 for convenience as all the numbers provided are powers of 2

Now we are set

$\log _{16}x + \log _{x}16 = \log_{512}x+\log _{x}512$

$\frac{\log_2x}{\log_216} + \frac{\log_216}{\log_2x} = \frac{\log_2x}{\log_2512} + \frac{\log_2512}{\log_2x}$

note that $\log_216 = \log_22^{4} = 4\log_22 = 4$ Similarly, $\log_2512 = 9$

$\frac{\log_2x}{4}+ \frac{4}{\log_2x} = \frac{\log_2x}{9} + \frac{9}{\log_2x}$

Multiplying throughout by $36\log_2x$ gives us

$9(\log_2x)^{2}+ 144 = 4(\log_2x)^{2} + 324$

let $y = \log_2x$

$9y^{2}+144 = 4y^{2}+324$

$5y^{2}=180$

$y = \pm\sqrt{36} = \pm 6$

$\log_2x = \pm 6$

$\log_2x = 6$ or $\log_2x = -6$

$x = 2^{6} = 64$ or $x = 2^{-6} = \frac{1}{64}$

the sum of all the possible values of x gives us $64+\frac{1}{64} = \frac{4097}{64}$

$4097+64 = 4161\Rightarrow$ the last 3 digits are $161$

$\log_{16} x + \log_{x} 16 = \log_{512} x + \log_{x} 512$

$\frac{\log x}{\log 16} + \frac{\log 16}{\log x} = \frac{\log x}{\log 512} + \frac{\log 512}{\log x}$

$\frac{\log x}{\log 2^{4}} + \frac{\log 2^{4}}{\log x} = \frac{\log x}{\log 2^{9}} + \frac{\log 2^{9}}{\log x}$

$\frac{\log x}{4 \log 2} + \frac{4 \log 2}{\log x} = \frac{\log x}{9 \log 2} + \frac{9 \log 2}{\log x}$

$\frac{\log x}{4 \log 2} - \frac{\log x}{9 \log 2} = \frac{5 \log 2}{\log x}$

$\frac{9 \log x}{36 \log 2} - \frac{4 \log x}{36 \log 2} = \frac{5 \log 2}{\log x}$

$\frac{5 \log x}{36 \log 2} = \frac{5 \log 2}{\log x}$

$\frac{\log x}{36 \log 2} = \frac{\log 2}{\log x}$

$(\log x)^{2} = 36(\log 2)^{2}$

$\log x = ±6 \log 2$

$\log x = \log 2^{-6}$ or $\log x = \log 2^{6}$

$\log x = \log \frac{1}{64}$ or $\log x = \log 64$

$x = \frac{1}{64}$ or $x = 64$

The sum of solutions is $\frac{1}{64} + 64 = \frac{4097}{64}$ . Let $a = 4097$ and $b = 64$ so that $a + b = 4161 ≡ 161$ mod $1000$ .

convert everything to $\log_2$ to get: $\frac{\log_2{x}}{\log_2{16}}+\frac{\log_2{16}}{\log_2{x}}=\frac{\log_2{x}}{\log_2{512}}+\frac{\log_2{512}}{\log_2{x}}$ $\frac{\log_2{x}}{4}+\frac{4}{\log_2{x}}=\frac{\log_2{x}}{9}+\frac{9}{\log_2{x}}$ Then let $y=\log_2{x}$ which means $x=2^{y}$ to get: $\frac{y}{4}+\frac{4}{y}=\frac{y}{9}+\frac{9}{y}$ $9y^2+4\times36=4y^2+9\times36$ $5y^2=5\times36$ $y^2=36$ $y=\pm6$ Therefore, $x=2^{6}$ or $x=2^{-6}$ which means the sum of these solutions is $2^{6}+2^{-6}=\frac{4097}{64}=\frac{a}{b}$ .

Therefore, $a+b=4097+64=4161$ and the last 3 digit are 161.

Note that 16 = 2^4 and 512 = 2^9. For convenience let X denote log x to the base 2. Given equation rewritten in terms of X becomes X/4 + 4/X = X/9 + 9/X. By inspection x = 1 (X=0) cannot be a solution. On multiplying every term by 36 X, we get 9X^2 + 144 = 4X^2 + 324, 5X^2 = 180, X^2 = 36, X = 6 or - 6, x = 64 or 1/64 Sum of the solutions is 4097/64 and 4097+ 64 being 4161. Hence answer is 161.

To begin with, note that for any real values of $p$ and $q$ , $\log_{q}p=\frac{\log_{m}p}{\log_{m}q}$ where $m$ is also any real number, which in this solution will be $10$ .

Therefore

$\log_{16}x+\log_{x}16=\log_{512}x+\log_{x}512 \Rightarrow$

$\Rightarrow \frac{\lg x}{\lg16}+\frac{\lg16}{\lg x}=\frac{\lg x}{\lg512}+\frac{\lg512}{\lg x}\Rightarrow \frac{\lg^{2}x+\lg^{2}16}{\lg16\lg x}=\frac{\lg^{2}x+\lg^{2}512}{\lg512\lg x}$

Now, using the fact that $\log m^{n}=n\log m$ , $16=2^{4}$ and $512=2^{9}$ , we may conclude $$

$\frac{\lg^{2}x+\lg^{2}16}{\lg16\lg x}=\frac{\lg^{2}x+\lg^{2}512}{\lg512\lg x} \Rightarrow \frac{\lg^{2}x+\lg^{2}2^{4}}{\lg2^{4}\lg x}=\frac{\lg^{2}x+\lg^{2}2^{9}}{\lg2^{9}\lg x}\Rightarrow$ $$ $\Rightarrow \frac{\lg^{2}x+16\lg^{2}2}{4\lg2\lg x}=\frac{\lg^{2}x+81\lg^{2}2}{9\lg2\lg x}\Rightarrow \frac{\lg^{2}x+16\lg^{2}2}{\lg^{2}x+81\lg^{2}2}=\frac{4}{9}\Rightarrow$

$\Rightarrow 9\lg^{2}x+144\lg^{2}2=4\lg^{2}x+324lg^{2}2\Rightarrow 5\lg^{2}x=180\lg^{2}2\Rightarrow$

$\Rightarrow lg^{2}x=36\lg^{2}2 \Rightarrow \lg^{2}x=\lg^{2}2^{6}\Rightarrow$

$\Rightarrow \lg^{2}x-\lg^{2}2^{6}=0\Rightarrow (\lg x-\lg2^{6})(\lg x+\lg2^{6})=0$ .

From this, we may derive $2$ separate cases:

Case I

$\lg x-\lg2^{6}=0\Rightarrow \lg x=\lg2^{6}\Rightarrow x_{1}=2^{6}$

Case II

$\lg x+\lg2^{6}=0\Rightarrow \lg x=-\lg2^{6}\Rightarrow \lg x=\lg2^{-6}\Rightarrow x_{2}=2^{-6}$

Thus, the sum of all the positive solutions equals

$x_{1}+x_{2}=2^{6}+\frac{1}{2^{6}}=\frac{2^{12}+1}{2^{6}}=\frac{4097}{64}$ .

Consequently, $\frac{a}{b}=\frac{4097}{64}\Rightarrow a+b=4161$ , meaning the answer is $161$ .

Let log2x=m. Then log16x= m/4 And logx16 is its reciprocal, i.e. 4/m. Similarly, log512x=m/9 and logx512=9/m. So the equation can be rewritten as m/4 + 4/m=m/9+9/m. 5m/36=5/m. m^2=36. m=+6, -6. 2^6=64=x. And 2^ (1/6) = 1/64=x. Adding both together, we obtain the sum of all possible values of x as 64+1/64, or 4097/64. Thus a=4097 and b=64. Therefore a + b = 4161. Last 3 digits of (a + b) are 161.