Exponential Equations and Inequalities and logarithms

Question

So this is one kind of task: $2\times 8^{x}-7\times4^{x}+7\times2^{x}-2=0$

And I don't get any of these tasks, so I am asking for some kind of literature with introduction about this and good explanation if you have to suggest. By the way we did exponential functions so it would be nice to give me some literature about exponential functions too...

By the way we started working with logarithms, basics things I do understand how to calculate: $\log _{2}4$ =?

, but I don't know how to calculate: $\log _{49} \sqrt[3]{7^{5}}$=? , so if you also have some nice pdf book with introduction and examples for logarithms too it will be nice.

Thank you.

Answer 1

The presence of summed terms "ruins" the prospects of apply logarithms directly. But you may sometimes be able to notice that an expression is a polynomial function of a single exponential expression. For your first problem, you can note that it can be rewritten as $$2(2^x)^3 -7(2^x)^2 + 7(2^x)-2=0$$ $$2u^3 -7u^2 + 7u-2=0$$ where $u=2^x$. If you can find the solutions (values of $u$) to this polynomial equation in $u$, then you can find the values of $x$ that solve the original equation by solving $u=2^x$ for $x$ (that is, $x=\log_2 u$) for each solution $u$.

For this particular polynomial, you probably notice immediately that $u=1$ is a solution since the coefficients add up to zero. This means you can factor the polynomial as $$(u-1)(\textrm{quadratic polynomial in }u)=0$$ which you can solve easily.

For the second problem, recall that $\log_b c=\frac{\log_a c}{\log_a b}$ for any choice of base $a$, and a convenient base to use is $a=7$ since both $b$ and $c$ are powers of $7$.

Answer 2

Hint: Write $$2\cdot (2^x)^3-7\cdot (2^x)^2-7\cdot 2^x-2=0$$ now let $$t=2^x$$ Can you finish? and it is $$\log_{49}7^{5/3}=\frac{\ln(7^{5/3})}{\ln(7^2)}=...$$

Answer 3

For the second question, there is this identity:

$$\log_b a = \frac{\log a}{\log b}$$

This is derived from: if you want to find out $\log_b a = ?$,

$$\begin{align*} \log_b a &= \ ?\\ a &= b^?\\ \log a &= \log \left(b^?\right) \tag{1}\\ &= \ ?\cdot\log b\\ ? &= \frac{\log a}{\log b} \end{align*}$$

For your example,

$$\begin{align*} \log_{49}\sqrt[3]{7^5} &= \frac{\log{\sqrt[3]{7^5}}}{\log 49}\\ &= \frac{\frac53\log 7}{2\log 7}\\ &= \frac56 \end{align*}$$

---

Note that, on line $(1)$, there is nothing special for taking base-$10$ logarithm. The base can be any base $c > 0$ and $c \ne 1$. The natural logarithm $\ln$ with base-$e$ is equally valid:

$$\begin{align*} ? &= \frac{\ln a}{\ln b}\\ &= \frac{\log_c a}{\log_c b} \end{align*}$$