Exponential equation: $5^{x+2}\cdot 2^{4-x} = 1000$

Question

I need help with this exponential equation: $5^{x+2}\ 2^{4-x} = 1000 $

We know that $ 1000 = 10^3$, so:

$$\ln(5^{x+2}\cdot2^{4-x}) = \ln10^3 \implies\ln(5^{x+2}) + \ln(2^{4-x}) = \ln10^3$$

In the next step I use that: $\ln(a^x) = x\ln(a)$

$$(x+2)\ln 5 + (4-x)\ln 2 = 3\ln 10$$

And I'm stuck here.

Answer 1

You can simplify a little more: $$\begin{align}(x+2)\ln 5 + (4-x)\ln2 &= 3\ln 10 \\ &=3\ln2 + 3 \ln 5 \\ (x-1)\ln5+(1-x)\ln2&=0.\end{align}$$ Can you finish it off from here?

Alternatively, you can do this without logs: $5^{x+2} \times 2^{4-x} = 2^3 5^3$, so $5^x 2^{-x} = \frac{5}{2}$, from where the answer should also be clear.

Answer 2

I see no reason to use logarithms here.

Yes, $1000= 10^3$ and $10= (2)(5)$ so $1000= (2^3)(5^3)$.

Your equation is $(2^{4- x})(5^{x+ 2})= (2^3)(5^3)$.

2 and 5 are prime numbers and prime factorization is unique so we must have 4- x= 3 and x+ 2= 3. That is two equations in one unknown but fortunately, x= 1 satisfies both.

Answer 3

$1000=5^{x+2}2^{4-x}=5^32^35^{x-1}2^{1-x}\iff 1=(5/2)^{x-1}\iff x=1$

Answer 4

Remember that $\ln 5$, $\ln2$ etc are just (real) numbers, so they can be used like real numbers, so: $$(x+2)\ln(5) + (4-x)\ln(2) = 3\ln(10)$$ expanding the brackets and factoring the $x$ s: $$x(\ln5 -\ln2) = 3\ln10-2\ln5-4\ln2$$ and combining the logs $$x\ln(5/2) = \ln1000-\ln25-\ln16 = \ln\left({1000\over 16\times25}\right)=\ln(2.5) $$ i.e. $$x\ln(2.5)=\ln(2.5)$$ and so $$x=1$$