Equation Involving Exponential Functions of Differing Bases I'm completely stumped here.
$$
2^{x+1} = 5^x
$$
If someone could explain how to solve for $x$ I'd be grateful.

Thanks.
Yet another way:
$$
\begin{align*}
2^{x+1} &amp;= 5^x\\
(x + 1)\ln 2 &amp;= x \ln 5\\
x\ln 2 + \ln 2 &amp;= x \ln 5\\
x \ln 2 - x \ln 5 &amp;= -\ln 2\\
x(\ln 2 - \ln 5) &amp; = -\ln 2\\
x &amp;= \frac{\ln2}{\ln5 - \ln 2}
\end{align*}
$$
Here's the step by step solution

$$2^{x+1} = 5^x$$

using the law of indices $$a^{b+c} =a^b a^c$$

$$log[2^x2^1] = log[5^x]$$

Taking log base 10 on both the sides and using one of the properties you get,

$$log (2^x) + log(2^1) = log(5^x)$$

using yet another property of log's we get

$$xlog2 + 1log2 = xlog5$$

x(0.3010) + 1(0.3010) = x(0.6989)

Now you know how to solve for x right?