Solving large exponents

Question

I have a problem that I have been working on and I am a little bit stuck on it.

We are given that $5^a = 1500$ and that $3^b = 3,333,333,333$

We are then given three different choices and told to select all that apply.

The choices are:

$ab \gt 3$

$a \gt b$

$2a \gt b$

I am a little bit confused how we solve for $a$ and $b$ given that the exponents are so large and we cannot use a calculator. Is there a quick way to find these values? Do we even need to know them to address the below relations (just the general magnitude)?

Tentatively, I am thinking this might involve some sort of prime factorization here as a first step? I am a little bit stuck, so any guidance, hints, or links to resources would be greatly appreciated.

Answer 1

You don't need to find $a$ and $b$ exactly to answer this question. You just need an estimate.

To estimate $a$, try computing $5^1, 5^2, 5^3, \ldots$ and use this to bound $a$ between two integers.

To estimate $b$, notice that $3^{b+1} = 3 \cdot 3^b = 3 \cdot 3333333333 = 9999999999 = 10^{10}-1$. Since $3^2 = 9 < 10$, we have $3^{20} = 9^{10} < 10^{10}-1 = 3^{b+1}$. What does this tell you about $b$?

Now, use the inequalities you found for $a$ and $b$ to see if you can conclusively prove or disprove each of those statements.

Answer 2

It is easy to see $a$ and $b$ are positive numbers. So check whether each inequality works or not:

$1) ab>3\quad\Rightarrow\quad5^{ab}>5^3\quad\Rightarrow\quad 1500^b>125$

The inequality holds because $1500^b>3^b=3333333333>125$

$2)a>b\quad\Rightarrow\quad5^a>5^b\quad\Rightarrow\quad1500>5^b$

It is false, because $5^b>3^b=3333333333$

$3)2a>b\quad\Rightarrow\quad5^{2a}>5^b\quad\Rightarrow\quad1500^2>5^b$

$1500^2=2250000$ and again $5^b>3^b=3333333333$. So it is false too.

Answer 3

Hijacking the comments that followed the question, and working in base $(10)$ logarithms, (accurate to 3 decimal places):

• $\log(5) \approx 0.699.$
• $\log(3) \approx 0.477.$
• $\log(2) \approx 0.301.$

Therefore:

$\displaystyle a(\log 5) = \log\left[\frac{3 \times 10^3}{2}\right] \implies a(0.699) \approx 3.176 \implies a \approx 4.544.$

$\displaystyle b(\log 3) \approx \log\left[\frac{10^{10}}{3}\right] \implies b(0.477) \approx 9.523 \implies b \approx 19.964.$

Therefore, the only correct choice is $ab > 3.$

Answer 4

Alternatively.

$ab > 3$:

Well notice that $5^a = 1500= 5\times 300 = 5\times 25\times 4 \times 3= 5^3 \times 12$ so $a > 3$.

And $3^b = 3,333,333,333= 3\times 1,111,111,111$ so $b > 1$. So $ab > 3\times 1 = 3$.

$a > b$:

$5^a = 1500 < 3,333,333,333 < 3^b < 5^b$ so $a < b$.

And $2a > 3$:

And $5^{2a} = 1500^2 < 2000^2 = 4,000,000 < 3,333,333,333 < 3^b < 5^b$ so $2a < b$.

Weird thing is none of these are even close.

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Alternatively we can put this to a common base logarithm.

$5^a = 1500$ so $\log_{10} 5^a = a \times \log_{10} 5$. And $\log_10 1500 = \log_{10} 5 + \log_{10}3 + 2$.

So $a = 3 + \frac {\log_{10} 3}{\log_{10} 5}$.

$3^b = 3,333,333,333 = \frac 13 (10^3 - 1)$

$\log_{10} 3^b = b \log_{10} 3$ and $\frac 13(10^{10}-1)$ is very close to $\frac 13 \times 10^{10} $ and $\log_{10}( \frac 13\times 10^{10})= 10 + \log_{10}\frac 13 = 10- \log_{10} 3$

and so $b$ is just teensy bit smaller than $\frac {10}{\log_{10} 3} - 1$.

Shouldn't be to difficult to convince ourselves that

$(3 + \frac {\log_{10} 3}{\log_{10} 5})(\frac {10}{\log_{10} 3} - 1) > 3$ and that $\frac {10}{\log_{10} 3} - 1 > 2(3 + \frac {\log_{10} 3}{\log_{10} 5}) > (3 + \frac {\log_{10} 3}{\log_{10} 5})$.