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.
 A: Yet another way:
$$
\begin{align*}
2^{x+1} &= 5^x\\
(x + 1)\ln 2 &= x \ln 5\\
x\ln 2 + \ln 2 &= x \ln 5\\
x \ln 2 - x \ln 5 &= -\ln 2\\
x(\ln 2 - \ln 5) & = -\ln 2\\
x &= \frac{\ln2}{\ln5 - \ln 2}
\end{align*}
$$
A: $2^{x+1} = 2^x \cdot 2^1$, so you want to solve $2 = 5^x / 2^x = (5/2)^x$. Hint: use logs.
A: First use the power law $a^{b+c}=a^b a^c$. Then isolate all $x$s on one side of the equals sign and use the law $\frac{a^c}{b^c} = (\frac{a}{b})^c$. Finally, logarithms.
A: HINT $\displaystyle\rm\quad \ 2^{\:X+1}\ =\ 5^{X}\ =\ 2^{\:\ell_2(5)\:X}\ \Rightarrow\ X+1\ =\ \ell_2(5)\ X\ \Rightarrow\ X\ =\ \ldots\:, \ $ for $\rm\ \ell_2\ =\ log_2$
NOTE $\rm\ \ \ $ We used $\rm\ \ Y\ =\ 2^{\:\ell_2(Y)}\:.\: $ Proof $\: $ Apply $\rm\:\ell_2\:,\:$ using $\:\ell_2(a^b) =\: b\ \ell_2(a),\ \ \ell_2(2) = 1\:.$
A: 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?
