I am learining about logarithm equations, and i can´t seem to understand how to solve such an equation, could someone help?
I must solve the equation/find $x$ for:
$$2^{2x} - 3\cdot2^x - 10=0$$
The final answer should be $x=\dfrac{\log5}{\log2}$
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Where is the equation sign? – zoli Mar 03 '15 at 15:15
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It´s an old textbook and it only says: Solve the equation(s) : 2^2x - 3 * 2^x - 10 = 0 – Jon Willy Larsen Mar 03 '15 at 15:17
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1A few $( \ ,\ )$ would be useful to make your equation clear. – Reveillark Mar 03 '15 at 15:19
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Please, edit. $$2^{2x}-2^x=10$$ would look much nicer. – zoli Mar 03 '15 at 15:24
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You have two equations that are not the same. The second is factored, so you just set each factor to zero. Neither gives the answer you seek. Also 2^2x should be read as $2^2x=4x$, but I suspect you mean $2^{2x}$. Please be careful with parentheses. – Ross Millikan Mar 03 '15 at 15:24
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If you put $y = 2^x$ then your equation for y is $y^2-3y-10=0$. Solve for y then solve $y = 2^x$ for x (if possible). – Paul Mar 03 '15 at 15:26
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Assuming you mean $2^{2x} - 3\times2^x - 10 = 0$, then since this is a quadratic in $2^x$ we can factorise as $(2^x+2)(2^x-5)=0$. Since $2^x>0$ $\forall x$, $2^x+2\neq0$, so we must have $2^x=5$. Taking logarithms and applying the power rule, $x\log2=\log5$, whence $x=\frac{\log5}{\log2}$ as required.
If you're having trouble seeing the factorisation, then try setting $y=2^x$ and solving the resulting quadratic for $y$, then using logarithms to obtain $x$.
Unochiii
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Could you explain how you factorised it to be: $$(2^x+2)(2^x−5)$$? – Jon Willy Larsen Mar 03 '15 at 19:17
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1One way to see that that I and others have mentioned is that if you let $y=2^x$, then $2^{2x}-3\times2^x-10=y^2-3y-10=0$, because $2^{2x} = 2^x\times2^x = (2^x)^2 = y^2$.
Then it's just a quadratic in $y$ - I assume you are familiar with its solution?
– Unochiii Mar 04 '15 at 00:50