I'm completely stumped here. $ 2^{x+1} = 5^x $ If someone could explain how to solve for $x$ I'd be grateful.
Thanks.
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} &= 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*} $
$2^{x+1} = 2^x \cdot 2^1$, so you want to solve $2 = 5^x / 2^x = (5/2)^x$. Hint: use logs.
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.
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 $\:\ell_2(a^b) =\: b\ \ell_2(a),\ \ \ell_2(2) = 1\:.$
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?