I‘m struggling right now with a equation. The equation is
$$\frac{2^{(5x)}}{7^{(x+2)}} = 10.$$
The solution should be $4.076$ but I don‘t know how to solve this equation.
I came to the conclusion that I can simplify the equation so it‘s $\frac{32^x}{7^{(x+2)}}$. My problem is that I don‘t know how I can get rid of the denominator.
Thank you for your help.
$$\dfrac{2^{5x}}{7^{x+2}}=10$$
$$\dfrac{32^x}{7^x49}=10$$
$$\left(\dfrac{32}7\right)^x=490$$
Can you take it from here?
You simplified $2^{5x}$ to $\left(2^5\right)^x=32^x$.
You can also simplify $7^{x+2}$ to $7^x\cdot7^2=49\cdot7^x$, using the exponent rule $a^{b+c}=a^b\cdot a^c$.
Thus, your equation simplifies to $$ \begin{align*} \frac{32^x}{49\cdot7^x}&=10\\ \frac{1}{49}\cdot\frac{32^x}{7^x}&=10\\ \frac{32^x}{7^x}&=490\\ \left(\frac{32}{7}\right)^x&=490\\ \ln\left(\left(\frac{32}{7}\right)^x\right)&=\ln(490)\\ x\ln\left(\frac{32}{7}\right)&=\ln(490)\\ x&=\frac{\ln(490)}{\ln\left(\frac{32}{7}\right)}\\ x&\approx4.076 \end{align*} $$
$$\frac{2^{(5x)}}{7^{(x+2)}}=10$$ $$\implies 2^{(5x)}=10\times7^{(x+2)}$$ Take logarithms of both sides and rearrange using the laws of logarithsms to find $x$, eg using $$\log a^b=b\log a$$.
