Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$3^x - 4^y = 5$,

$3^{x+1} + 4^y = 23$

These are the two equations to be solved correct to 3 significant figures.

I did this: $\lg 3^x - \lg4^y=\lg5$

$ x\lg3-y\lg4=lg5 $


Is this the right way?

I don't think so, cause I continued and the answer was wrong! :'( Help!

Book answer: $x=1.77$ and $y=0.500$

share|cite|improve this question
Nope: $\log(a-b) \not\equiv \log(a) - \log(b)$. (which is an error similar to another false assumption you made in a previously posted question!) -- I suggest that you read back over your log laws and ensure that you are familiar with them first. – FH93 Feb 11 '14 at 21:23
up vote 1 down vote accepted

Let $X=3^x$ and $Y=4^y$. Noting that $3^{x+1} = 3\cdot 3^x$, our system reduces to the pair of linear equations:

$\begin{eqnarray*} X-Y &=& 5 & (1)\\ 3X+Y &=& 23 &(2) \end{eqnarray*}$

Note $(2)+(1): 4X = 28$ so $X=7$; and upon back substituting this into $(1)$, we obtain $Y=2$ (you can verify that $(X,Y)=(7,2)$ satisfy $(2)$)

Hence, undoing our transformation of variables:

$X=3^x = 7 \implies x=\log_3 (7)$

$Y=4^y = 2 \implies y=\log_4(2) = \log_4(4^{1/2}) = 1/2$

share|cite|improve this answer
It is not pilot to restate other solutions – Babak Miraftab Feb 11 '14 at 21:35
I apologise, I hadn't realised that an answer had been posted -- internet isn't great here. – FH93 Feb 11 '14 at 21:58

Another solution is that you can add two equations and so $3^x+3^{x+1}=28$. Thus, we have $3^x(1+3)=28$ and it implies that $3^x=7$ and so $x=log_37$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.