Working with the equation: $25^{\log_{10}(x)}=5+4x^{\log_{10}(5)}$

Question

The answer should be $10$, but I don't know how to get it. I've tried many different ways and I still can't get it. I tried this : $$25^{\log_{10}(x)}=5+4x^{\log_{10}(5)}\implies \log_{10}(25^{\log_{10}(x)} - 5) = \log_{10}(4x) \cdot \log_{10}(5).$$

Answer 1

You have $$25^{\log_{10}(x)} = 5^{2 \log_{10}(x)} = e^{2 \frac{\ln(x)}{\ln(10)} \ln(5)} = x^{2 \log_{10}(5)}$$

So, let's define $X = x^{\log_{10}(5)}$. Your equation becomes $$X^2 = 5 + 4X$$

i.e. $X=5$ or $X=-1$. Of course $X=-1$ is impossible because by definition $X>0$. So $X=5$, i.e. $$x^{\log_{10}(5)} = 5$$

You obtain finally $$x = 5^{\frac{1}{\log_{10}(5)}} = 5^{\frac{\ln(10)}{\ln(5)}} = e^{\frac{\ln(10)}{\ln(5)} \ln(5)} = e^{\ln(10)} = 10$$

Answer 2

Hint: make a change: $x=10^t$.