# 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.

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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*}

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More natural, I think, because of the immediate reduction to a linear problem. –  André Nicolas Sep 5 '11 at 22:07
The common logarithm can be used instead of the natural one, of course, and the solution's still the same. Alternatively, one might write the solution as $$\frac1{\log_2 5-1}$$ –  Ｊ. Ｍ. Sep 6 '11 at 1:31

$2^{x+1} = 2^x \cdot 2^1$, so you want to solve $2 = 5^x / 2^x = (5/2)^x$. Hint: use logs.

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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.

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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 $\:\ell_2(a^b) =\: b\ \ell_2(a),\ \ \ell_2(2) = 1\:.$

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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?

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alok, Good attempt, but I am not sure the OP wants a numerical answer. It may be more helpful if get the exact answer first, and then plug in the numerical values. –  Srivatsan Sep 12 '11 at 1:11