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 $x=\dfrac{\log5}{\log2}$
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 $x=\dfrac{\log5}{\log2}$
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 $x$.