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"Log"ing Problem!

Posted: Thu Apr 05, 2018 4:50 pm
by Tasnood
Let $a>1$ and $x>1$ satisfy $log_a(log_a(log_a2)+log_a24-128)=128$ and $log_a(log_ax)=256$.
Find the remainder when $x$ is divided by $1000$

Re: "Log"ing Problem!

Posted: Thu Apr 05, 2018 4:56 pm
by Tasnood
$\Rightarrow log_a(log_a2^{24})=a^{128}+128$
$\Rightarrow log_a2^{24}=a^{a^{128}+128}$ [According to the basis of logarithm]
$\Rightarrow 2^{24}=a^{a^{a^{128}+128}}$
$\Rightarrow a^{a^{128}.a^{a^{128}}}=2^{24}$
$\Rightarrow y^y=2^{24}$ [Setting $a^{a^{128}}=y$]
$\Rightarrow y^y=8^8$ And we get: $y=8$

We get: $y=a^{a^{128}}=2^3$. Let $z=a^{128}=2^l$ So, $y=a^z$

If $z=128$, $y=a^z=a^{128}=z=128>8$; Then $z\ne128$.
If $z=64=2^6$, $y=a^z=a^{64}=z^{\frac{1}{2}}=\sqrt{64}=8$. So we were right in imagination and $z=64=2^6=a^{128}=a^{2^7}$

Now let's come to the second given equation.
$\Rightarrow log_ax=a^{256}$
$\Rightarrow x=a^{a^{256}}=a^{a^{128+128}}=a^{a^{128}.a^{128}}=a^{z^2}$
We know: $z=2^6 \Rightarrow z^2=(2^6)^2=2^{12}$

So, $x=a^(z^2)=a^{2^12}=a^{{2^7}.{2^5}}=(a^{2^7})^{2^5}=(a^{128})^{2^5}=z^{32}=64^{32}=2^{192}$

Now the question is simple. What is the remainder when we divide $2^{192}$ by $1000$?
It is easy to say:$2^{192}\equiv 0$ (mod $8$)

GCD$(2,125)=1$. So, $\varphi (125)=\varphi (5^3)=5^3-5^2=100$

Implying Euler's Theorem, we get: $2^{\varphi(125)}\equiv1$ (mod $125$)$\Rightarrow 2^{100}\equiv1$ (mod $125$)
Some calculation will give the answer. I think I am correct :oops: