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

  • 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.

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

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


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

Unfortunalety, this is where I'm stuck.

With my kindest regards!

  • 1
    $\begingroup$ HInt: Factor the difference of squares. $\endgroup$ – Ethan Bolker Apr 24 '18 at 20:25

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


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


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.


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:

$\log_b k = c \implies$

$b^c = k \implies$

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

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

$Mc = N\implies$

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

  • $\begingroup$ Thanks for the first sentence. Thanks too for the logarithm lesson. I the OP will take advantage of it. I think it unlikely that others will find their way to this answer. $\endgroup$ – Ethan Bolker Apr 24 '18 at 21:19
  • $\begingroup$ I think the OP is responding to an earlier question she/he had. math.stackexchange.com/questions/2752147/find-the-value-of-ab. Being asked to solve if $5^a =16$ and $8^b = 25$ what is $ab$. Everyone assumed the straightforward thing was to do simple logarithms, but it came out the OP didnt know logarithms and the teacher was expecting cleverness. This problem is similar although her it's more clear (if one knows logarithms) that logarithms will be difficult, but cleverness ... well, once you see the clever trick will be ... clever. $\endgroup$ – fleablood Apr 24 '18 at 21:58

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

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

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


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

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

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

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


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