Equation involving exponents

$\begin{aligned} 4^x+10^x & = 25^x & \small \blue{\text{Divide both sides by }10^x} \\ \left(\frac 25\right)^x + 1 & = \left(\frac 52\right)^x & \small \blue{\text{Multioly both sides by }\left(\frac 25\right)^x} \\ \left(\frac 25\right)^{2x} + \left(\frac 25\right)^x - 1 & = 0 & \small \blue{\text{and add both sides by }-1} \\ \implies \left(\frac 25\right)^x & = \frac {-1 + \sqrt 5}2 & \small \blue{\text{Since }\left(\frac 25\right)^x > 0} \\ x (\ln 2 - \ln 5) & = \ln (\sqrt 5 - 1) - \ln 2 \end{aligned}$

$\begin{aligned} \implies x & = \frac {\ln 2 - \ln (\sqrt 5-1)}{\ln 5-\ln 2} \approx \boxed{0.525} \end{aligned}$

$\begin{aligned} \big(2^x + \tfrac12\cdot5^x\big)^2 & = \; \tfrac54 \cdot25^x \\ 2^x + \tfrac125^x & = \; \tfrac12\sqrt{5}\cdot 5^x \\ 2^{x+1} & = \; (\sqrt{5}-1)5^x \\ (x+1) \ln 2 & = \; \ln(\sqrt{5}-1) + x\ln 5 \\ x & = \; \boxed{\frac{\ln2 - \ln(\sqrt{5}-1)}{\ln5 - \ln2}} \end{aligned}$

Divide the original equation by 4^x the resulting equation will look like

1 + (10^x/4^x) = 25^x/4^x =

1 + (5/2)^x = ((5/2)^x)^2

Let (5/2)^x = t

The equation becomes

1+t = t^2 or t^2 - t -1 = 0 or t = (1+/- sqrt(5))/2

So (5/2)^x = 1.62

x * ln(2.5) = ln(1.62) so x = 0.52517

Look for root to $25^x - 10^x - 4^x$ using Excel.

$25^1 - 10^1 - 4^1 = 11$

$25^0 - 10^0 - 4^0 = -1$

$25^{0.5} - 10^{0.5} - 4^{0.5} = -0.16228,$

... (skipping past computations for $x=0.6, 0.55. 0.53$ , and $0.52$ )

$25^{0.525} - 10^{0.525} - 4^{0.525} = -0.00119$

$25^{0.526} - 10^{0.526} - 4^{0.526} = 0.005685$

So, the root is between $0.525$ and $0.526$ , and looks to be closer to the former.