How to solve for $x$ in $2^x=5^{x+2}$

Question

So I was looking through the homepage of Youtube when I came across this video by Blackpenredpen which asked for the value of $x$ in $$2^x=5^{x+2}$$which I thought that I would be able to solve. Here is my attempt at solving the aforementioned equation:$$2^x=5^{x+2}$$$$x\ln2=(x+2)\ln5$$$$x=\frac{(x+2)\ln5}{\ln2}$$$$\frac{x}{x+2}=\ln(5)-\ln(2)$$$$\frac{x}{\ln(5)-\ln(2)}=x+2$$$$\frac{x}{\ln(5)=\ln(2)}-2=x$$$$\frac{\ln(5)-\ln(2)}{x}-0.5=\frac{1}{x}$$$$\ln(5)-\ln(2)-0.5x=1$$$$\ln(5)-\ln(2)=0.5x+1$$$$\ln(5)-\ln(2)-1=0.5x$$$$2\ln(5)-2\ln(2)-2=x$$$$\mathbf{\text{My question}}$$

---

Is the solution that I arrived at correct, or what could I do to attain the correct solution or attain it more easily?

Answer 1

Your answer is not correct. The correct answer proceeds as follow: $$ 2^x=5^{x+2}$$ $$x\ln2=(x+2)\ln5$$ $$x\ln2=x\ln5+2\ln5$$ $$x\ln2-x\ln5=\ln25$$ $$x(\ln2-\ln5)=\ln25$$ $$x=\frac{\ln25}{\ln2-\ln5}$$ $$x =\frac{\ln 25}{\ln \frac 25}$$

Answer 2

As noticed there is an issue with an intermediate step, more easily we can proceed as follows

$$2^x=5^{x+2} \iff 2^x=5^x\cdot5^2 \iff \left(\frac25\right)^x=25 \iff x\ln \frac 25 =\ln 25 \iff x =\frac{\ln 25}{\ln \frac 25}$$

Answer 3

Is your solution correct?

Unfortunately not. You can check it by entering the result in the original equation. You made a mistake from $$\frac{x}{\ln(5)-\ln(2)}-2=x$$ to $$\frac{\ln(5)-\ln(2)}{x}-0.5=\frac{1}{x}$$

How to attain a solution more easily?

Try to get rid of the parentheses in $$x\ln 2 = (x+2)\ln 5$$

Answer 4

There's a problem between your line 6 and 7. Here's a faster way. $$2^x=5^{x+2}\iff x\ln(2)=(x+2)\ln(5)\iff x\ln(2)-x\ln(5)=2\ln(5)\iff x=\frac{2\ln(5)}{\ln(2)-\ln(5)}=-\frac{2\ln(5)}{\ln(\frac52)}$$