1. ## Exponential Contest Problem

Hello everyone,

Could someone pleae check my solution for the following problem?

Thank you!

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1. Let $\displaystyle a$ and $\displaystyle b$ be real numbers with $\displaystyle a > 1$ and $\displaystyle b > 0$. If $\displaystyle ab = a^b$ and $\displaystyle \frac{a}{b} = a^{3b}$, determine the value of $\displaystyle a$.

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When the first equation is multiplied by a:

$\displaystyle b = a^{b - 1}$ --(3)

After substitution of (3) in the second equation:

$\displaystyle \frac{a}{a^{b - 1}} = a^{3b}$

$\displaystyle a^{2 - b} = a^{3b}$

Therefore:

$\displaystyle 2 - b = 3b \Rightarrow b = 1/2$

After substitution of b into (1):

$\displaystyle 0.5a = \sqrt{a}$

$\displaystyle 0.25a^2 - a = 0$

$\displaystyle a(0.25a - 1) = 0 \Longrightarrow a = 0, 4$

2. Originally Posted by scherz0
Hello everyone,

Could someone pleae check my solution for the following problem?

Thank you!

---

1. Let $\displaystyle a$ and $\displaystyle b$ be real numbers with $\displaystyle a > 1$ and $\displaystyle b > 0$. If $\displaystyle ab = a^b$ and $\displaystyle \frac{a}{b} = a^{3b}$, determine the value of $\displaystyle a$.

---

When the first equation is multiplied by a:

$\displaystyle b = a^{b - 1}$ --(3)

After substitution of (3) in the second equation:

$\displaystyle \frac{a}{a^{b - 1}} = a^{3b}$

$\displaystyle a^{2 - b} = a^{3b}$

Therefore:

$\displaystyle 2 - b = 3b \Rightarrow b = 1/2$

After substitution of b into (1):

$\displaystyle 0.5a = \sqrt{a}$

$\displaystyle 0.25a^2 - a = 0$

$\displaystyle a(0.25a - 1) = 0 \Longrightarrow a = 0, 4$
yes, you are right. Also,

since, a > 1

so, a = 4 and b = 1/2 is the solution

3. Thank you for your reply, shyam and for noting that $\displaystyle a > 1$ as I had forgotten!