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Solve the inequality $25^x>3000$.

Note: No computer programs or calculators allowed.

I tried factoring the two, I ended up with $5^{2z}>2^3\cdot3\cdot5^3$, which doesn't seem to help.

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    Hint: $\log$ each side. – 2017-02-23
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    @MathematicsStudent1122 $x>\dfrac{\log3000}{\log(25)}$? Can I simplify this further? – 2017-02-23
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    Well, is $x$ an integer or a real? – 2017-02-23
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    $x$ must be greater than a value a little under $2.5$, based on $5^3\cdot 24$ – 2017-02-23
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    @S.C.B. real $\!$ – 2017-02-23
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    Take log base 5 and 3000=25x 120 ~=5^5 gives a very close estimate of x >2.5. Just a smidgen less maybe. – 2017-02-23
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    "Which doesn't seem to help". It actually helps a lot. Take log base five of each side and you get 2z > 3+log_5 24 approx 5. – 2017-02-23

5 Answers 5

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$$25^{x} > 3000$$

$$5^{2x} > 3 \times 10^3 = 3 \times 2^3 \times 5^3 $$

$$5^{2x-3} > 24$$

$$2x-3 > \log_5 24$$

$$x > \frac{\log_5 24 + 3}{2}$$

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    And log_5 24 is just a smidgen under 2, if we want some ball park figure. – 2017-02-23
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$25^x > 3000$

$\log_{25}25^x>\log_{25}3000$

$x >\log_{25}3000 $

Now $3000=25*120=25*25*4.8$ so $\log_{25}3000=2+\log_{25}4.8$ and $4.8 < 5$ so $\log_{25}4.8 < 1/2$ and $\log_{25}3000\approx 2.45$ or so.

So $x>2.45 $ or so.

Check.

If $x > 2.5$ then $25^x > 25^{2.5}=25^2\sqrt {25}=25*25*5=25*125=5*625=3125$

Close. $2.45$ might be closer.

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If we wan't to get technical. Let $-\epsilon =\log_{25} 24/25$ which is something negative be very close to zero. $\log_{25}3000=\log 25*5*24=\log 25*5*25*\frac {24}{25}=2.5 -\epsilon $. We can tear our hair out estimating $\epsilon$ but... it is very small.

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Or to continue what you did.

$5^{2z}>2^3*3*5^3=24*5^3\approx 5^5$

So $z \sim > 2.5$.

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Assume $25^x >3000$. Divide this inequality by $25$, getting

$25^{x-1} > 120$. That is $5^{2x-2}> 120$. As 120 is close to 125, choosing $2x-2\geq 3$ will definitely be enough. That is $x> 2.5$. For some slightly smaller $x$ will also be tru as we replaced 120 by 125.

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25^X greater than 3000 Taking log X log (25) greater than log (3000)

X grater than log (3000)/log 25

Now I think you could probably do it by using a log table. It will still be complicated but u don't have to use a calc.

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    There are many tricks to make guesses without tables. 3000 =3x10^3. And 25=5^2. So the answer is $(3+\log 3)/(2\log 5) \approx 3.5/2*\frac 23 \approx 2.4$. – 2017-02-23
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$$25^x>3000 \implies 5^{2x}>3\cdot10^3 \implies 5^{2x} > 3\cdot5^3\cdot2^3$$ $$\frac{5^{2x}}{5^3} > 3\cdot2^3 \implies 5^{2x-3} > 3\cdot2^3$$ $$x>\frac{(\log_5(24)+3)}{2}$$ $$x>2.4873179343530822$$



$25^{2.487317934353082\color{#f00}{2}} = 2999.9999999999997723697$ $25^{2.487317934353082\color{#f00}{3}} = 3000.0000000000007380324$