exponential equation

$\hspace{-16}Solve the equation \\\\ $\mathbf{64^x-27=343^{x-1}+\frac{3}{7}.28^x}$

5 Answers

We rewrite it as 4^{3x} - 3^{3} = 7^{3(x-1)} + 3.x^{x}.7^{x-1}

From here, we could set 7^{x-1} = a and 4^{x} = a and Probably solve.

I think you meant 7^{x-1}=a and 4^{x}=b.

1708

man111 singh ·2011-11-20 04:34:54

Yes Vivek and Sambit.

But, that substitution doesnt help.
I dont think there's any integer solution for this one.

Hmm, I see that too. Wolfram alpha says the two roots are

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