mathematics

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# mathematics

Submitted by adisa on Fri, 04/01/2011 - 14:45

Forums:

i need a solution to this question : 5^x equals 10(2^x)-15. I kow the answer is one but i dont know the steps in solving it.

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## Re: mathematics

The type of your equation is in general impossible to solve exactly; it contains two different exponential term and a constant term. You cannot join the exponential terms to one exponential term e.g. through a division without that the constant term forms a new exponential term [but if the original equation lacks the constant term, you would get the simple eq. (5/2)^x = 10].

Your example is a very special case, since it has two integer solutions which may be found by experimenting. You however see the difficulty of the equation type if you alter your example a little. Try to solve e.g.

5^x = 11*2^x-15.

It has two irrational roots near 1 and 2, but one cannot express them as exact expressions.

Best regards,

Jussi

## Re: mathematics

Hi adisa,

what pahio explained you is correct. In general we don't know an algebraic technique to solve a transcendent Eq. having the general form

a^x + b = c.d^x.

However, in your very particular example

5^x + 15 = 10.2^x, or 5^x + 15 - 10.2^x = 0 (1)

a simple inspection reveals us that x = 1 and x = 2 are solutions. There may be another root? No, why? Because for x > 2, always

5^x + 15 > 10.2^x.

It is a problem of monotony, essentially because 5^x > 2^x, e.g. for x = 3, we see that the LHS of (1) is 60. So that, increasing 'x' the positive deviation from '0' will be every time bigger.

The moral about this problem is you must check whether an elemental solution exists or not.

## Re: mathematics

It should be enough to assert the uniqueness of the solution $x_{0} =1$ by studying the behaviour of the smooth function $g(x) = 5^{x}- 10 2^{x} +15$.