Odd Base

Algebra Level 5

Let a = 3 2 + log b ( 11 5 25 + 2 5 25 i ) + i π log b e \large{a = \frac{3}{2} + \log_b\left({\dfrac{11\sqrt{5}}{25} + \dfrac{2\sqrt{5}}{25}i}\right) + i\pi \log_b{e}} and b a = z , b^a = z, where z z is of the form A + B i . A + Bi. If z 2 + ( z ) + ( z ) = 112 , |z|^2 + \Re(z) + \Im(z) = 112, what is the value of b ? b?


The answer is 5.

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1 solution

Guilherme Niedu
Jan 6, 2017

Powering from base b b in both sides:

b a = b 3 / 2 ( 11 5 25 + i 2 5 25 ) ( b l o g b e ) i π \large b^a = b^{3/2} \cdot (\frac{11\sqrt{5}}{25} + i \frac{2\sqrt{5}}{25} ) \cdot (b^{log_be})^{i\pi}

z = b 3 / 2 e a t a n ( 2 / 11 ) e i π \large z = b^{3/2} \cdot e^{atan(2/11)} \cdot e^{i \pi}

Then:

z 2 = b 3 \large |z|^2 = b^3

R e ( z ) = b 3 / 2 11 5 25 \large Re(z) = -b^{3/2}\frac{11\sqrt{5}}{25}

I m ( z ) = b 3 / 2 2 5 25 \large Im(z) = -b^{3/2}\frac{2\sqrt{5}}{25}

Summing up and multiplyting both sides by 25 25 :

25 b 3 13 5 b 3 / 2 2800 = 0 \large 25b^3 - 13\sqrt{5} b^{3/2} - 2800 = 0

Solving for b 3 / 2 b^{3/2} :

b 3 / 2 = 13 5 ± 237 5 50 \large b^{3/2} = \frac{13 \sqrt{5} \pm 237\sqrt{5}}{50}

b 3 / 2 = 5 5 \large b^{3/2} = 5\sqrt{5}

b 3 = 125 \large b^3 = 125

b=5 \color{#3D99F6} \fbox{b=5}

Isn't this with the assumption that b is real, which need not be the case?

Aditya Dhawan - 4 years, 4 months ago

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Isn't the logarithm function defined for only positive bases (excluding 1)?

Tapas Mazumdar - 4 years, 4 months ago

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That is true only for the ' Real Logarithmic Function '. The complex analogue deals with all complex bases. Explicitly, one may define log a b = ln b ln a , { a , b C , ln a 0 } \log _{ a }{ b } =\quad \frac { \ln { b } }{ \ln { a } } ,\quad \{ \quad a,b\quad \in \quad C,\quad \ln { a\neq 0 } \quad \} \quad

Aditya Dhawan - 4 years, 4 months ago

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