High school math problem

Algebra Level 2

$\ln(a^{1+2+3+\dots+n})=3n(n+1)$

Find value of the constant $a$ satisfying the equation above.

e^{10}

\sqrt{e}+1

\sin \sqrt{3}

e^{6}

2 solutions

Chew-Seong Cheong
Mar 9, 2020

Similar solution with @Sam sam 's.

$\begin{aligned} \ln (a^{1+2+3+\cdots + n}) & = 3n(n+1) \\ \ln (a^{\frac {n(n+1)}2}) & = 3n(n+1) \\ \frac {n(n+1)\ln a}2 & = 3n(n+1) \\ \ln a & = 6 \\ \implies a & = \boxed {e^6} \end{aligned}$

Sam Sam
Mar 8, 2020

first using rules of logs, we have

$\ln(a^{1+2+3+...+n})=\ln(a)(1+2+3+...+n)$

🠢 $\ln(a)=\frac{(3n(n+1))}{(1+2+3+...+n)}$

since $e^{\ln(a)}=a$

we can isolate $a$ by rewriting as follows

🠢 $a=e^{\frac{(3n(n+1))}{(1+2+3+...+n)}}$

note that the arithmetic sum can be expressed as $\frac{n(n+1)}{2}$

so we can rewrite $\frac{3n(n+1)}{(1+2+3+...+n)}$ = $\frac{2\times3(n(n+1))}{(n(n+1))} = 6$

🠢 $a=e^{6}$

@Sam sam , you need to put a backslash "\" in front of all math function including \ln $\ln$ , \sum $\sum$ , \sin $\sin$ , \cos $\cos$ . They should not be in italic.

Chew-Seong Cheong - 1 year, 3 months ago

Thanks, ill fix it

Sam sam - 1 year, 3 months ago

High school math problem

2 solutions

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