What is the value of $4^{a+b}$?

Question

$$x^{a^2-b^2} = 5, x^{a-b}=2$$

• What is the value of $4^{a+b}$?

I'll try to solve this question by using logarithm even thought I don't have any knowledge regarding to what logarithm actually means.

$x^{a^2-b^2} = 5$

$$\frac{\log (5)}{\log (x)} = a^2-b^2 \tag{1}$$

$x^{a-b}=2$

$$\frac{\log (2)}{\log (x)} = a+b \tag{2}$$

Unfortunalety, this is where I'm stuck.

With my kindest regards!

Answer 1

Since $a^2-b^2 = (a-b)(a+b)$, then $5 = x^{a^2-b^2}=(x^{a-b})^{a+b} = 2^{a+b}$.

So $4^{a+b} = (2^{a+b})(2^{a+b}) = 5 \cdot 5 = 25$.

Answer 2

$$x^{(a^2-b^2)} = x^{(a+b)(a-b)} = (x^{a-b})^{(a+b)} = 2^{a+b}$$ Then: $2^{a+b}=5$ hence $4^{a+b} = (2^{a+b})^2 = 5^2=25$

Answer 3

Don't use logarithms if you don't know what they mean! (Learn what they mean and then use them!)

Use $a^2 - b^2 = (a+b)(a-b)$.

$4^{a+b} = 2^{2(a+b)} = (x^{a-b})^{2(a+b)} = x^{2(a+b)(a-b)} = x^{2(a^2 - b^2)} = (x^{a^2 - b^2})^2=5^2 = 25$.

.... or ....

$5= x^{a^2 - b^2} = x^{(a+b)(a-b)}=(x^{a-b})^{a+b} = 2^{a+b}$

so $5^2 = (2^{a+b})^2 = (2^2)^{a+b} = 4^{a+b}$.

By the way. Your teacher is weird.

But I like him or her.

====

By the way. What logarithm means is a method of equating things be what power you must raise bases to.

$b^c = k \iff \log_b k = c$. that is the definition of a logarithm. If you can get to $c$ by raising $b$ to a power, $k$, then that power $k$ is called the logarithm of $c$ in base $b$.

$\log_b k$ means: the power you must raise $b$ to, in order to get $k$.[1]

So to do this with logs:

$x^{a^2 - b^2} = 5$. So $\log_x 5 = a^2 - b^2$.

And $x^{a-b} = 2$. So $\log_x 2 = a-b$.

Now $a^2 - b^2 = (a+b)(a-b)$. So

$\log_x 5 = (a+b) \log_x 2$.

Now here is one very basic identity:

If $b^k = c$ then $b^{mk} = c^m$. So $\log_b c^m = mk = m*\log_b c$.

So $(a+b)\log_x 2 = \log_x 2^{a+b}$.

So $\log_x 5 = \log_x 2^{a+b}$

Oh, here is another basic identity. $\log_b M = \log_b N \iff N=M$. This is obvious. $\log_b M = k\implies b^k = M$, and $\log_b N = k\implies b^k = N$. So $M = b^k = N$.

So if $\log_x 5 = \log_x 2^{a+b}=k$ then $2^{a+b} = 5$.

So $4^{a+b} = (2^{a+b})^2 = 5^2 = 25$.

..... which is a lot to throw at you all at once...

but trust us. It's very useful once you get used to it. You'll be able to solve things directly rather than just looking for tricks.

[1]

Some basic rules:

$b^{\log_b k} = b^{\text{the power we must raise b to in order to get k}} = k$.

If $m = b^k$ and $n = b^j$ then $nm = b^kb^j = b^{k+j}$ so $\log_b nm = k+j = \log_b n + \log_b m$.

If $m = b^k$ and $m^j = (b^k)^j = b^{kj}$ then $\log_b m^j = kj = j\log_b m$.

If $\log_b a = \log_b c$ then $a = c$. ... Just do it.... $a = b^{\log_b a} =b^{\log_b c } = c$.

And if $\log_{10} b = M$ and $\log_{10} k = M$. then:

$\log_b k = c \implies$

$b^c = k \implies$

$(10^M)^c = 10^N\implies$

$10^{Mc} = 10^N\implies$

$Mc = N\implies$

$\log_b k = c = \frac NM = \frac {\log_{10}k}{\log_{10} b}$

Answer 4

Here is a solution by using logarithmic properties, if you are interested.

$log_x 5=a^2-b^2$, $log_x 2=a-b$,

$\frac{log_x 5}{log_x 2}=\frac{(a+b)(a-b)}{(a-b)}$, $log_2 5=a+b$

$4^{a+b}=4^{log_2 5}=2^{2log_2 5}=2^{log_2 25}=25$

Answer 5

I would not do this using logarithms. But as you have already started down that path.

$a+b = \frac {\log 2}{\log x}\\ a^2-b^2 = \frac {\log 5}{\log x}\\ (a+b)(a-b) = \frac {\log 2}{\log x}\\ (a-b)\frac {\log 2}{\log x} = \frac {\log 5}{\log x}\\ (a-b) = \frac {\log 5}{\log 2} = \log_2 5$

$4^{a-b} = 4^{\log_2 5}\\ 2^{2\log_2 5}\\ 2^{\log_2 5^2}\\ 25$