2018 in Algebra (3)

Algebra Level 2

$a\log_{10}2+b\log_{10}3+c\log_{10}5+d\log_{10}7= 2018.$

$a,b,c$ and $d$ are integers that satisfies the equation above. Find $a+b+c+d$ .

1009 4036 2017 2018 4034

2 solutions

Mr. India
Mar 30, 2019

If $\log_{c}{b}=a$ then $c^a=b$

$\log_{m}{n}+\log_{m}{l}=\log_{m}{(n.l)}$

$a\log_{c}{b}=\log_{c}{b^a}$

So, given equation can be written as

$\log_{10}{(2^a.3^b.5^c.7^d)}=2018$

Removing logarithm

$10^{2018}=2^a.3^b.5^c.7^d$

$2^{2018}.5^{2018}=2^a.3^b.5^c.7^d$

Equating the exponents we get

$a=2018,b=0,c=2018,d=0$

So, $a+b+c+d=\boxed{4036}$

Steven Perkins
Mar 7, 2018

If you combine the logarithms, then take the anti-log of both sides you can see that the solution in integers (since 2 and 5 are factors of 10) must be:

$2^a*5^c = 10^{2018}$

And thus a=c=2018.

Did the same, but only thing that nags is the question : it states that a, b, c, d are positive reals. But now b=d=0.

Peter van der Linden - 3 years, 3 months ago

Hmmm. I was thinking it said "integers".

Now it says real numbers and there are multiple solutions and the sum is not unique.

For example, it can be solved with any three variables zero and the answer is different in all 4 cases.

Steven Perkins - 3 years, 3 months ago

2018 in Algebra (3)

2 solutions

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