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

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

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 $

$x\lg3-y=\frac{\lg5}{\lg4}$

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 $y=0.500$

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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 at 21:23
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2 Answers 2

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:

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

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

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It is not pilot to restate other solutions –  Babak Miraftab Feb 11 at 21:35
    
I apologise, I hadn't realised that an answer had been posted -- internet isn't great here. –  FH93 Feb 11 at 21:58
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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 $x=log_37$.

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