Exponential equation with logs

Dank2 · Sep 5, 2019

How can i continue from here, answer is x=0,-2

Handwritten derivation on squared paper solving log(25^(x+1) + 4^(x+1) - 19·10^x) - log(10^(x+1)) = 0, with steps expanding powers to 5^x·5^x·25 + 2^x·2^x·... and a crossed-out term.

mfb · Sep 5, 2019

You lost a factor 4 for 2^x 2^x.

I would make some common factors of 10^x and from there on say that you can see the solution. There is no nice way to solve this as far as I can see.

Dank2 · Sep 5, 2019

mfb said:

You lost a factor 4 for 2^x 2^x.

I would make some common factors of 10^x and from there on say that you can see the solution. There is no nice way to solve this as far as I can see.

You mean write down 25 as 10^log₁₀25?

Handwritten working on squared paper: the expression [10^(log10 25)]^x times 25 plus [10^(log10 4)]^x times 4 minus 19 times 10^x, followed by the line 10^(x log10 25) times 25 plus 10^(x(log10 4)) times 4 minus 19 times 10^x equals 0, with two crossed-out marks beneath.

Dank2 · Sep 5, 2019

dont mind the t=0, since it cannot be.

Handwritten working on squared paper: the equation (10^(log10 25))^x · 25 + (10^(log10 4))^x · 4 − 19·10^x = 0, rewritten as 10^(x log10 25)·25 + 10^(x(log10 4))·4 − 19·10^x = 0, then with 10^x = t, becoming 25t^(log10 25) + 4t^(log10 4) − 19t = 0, factored as t(25t^(log10 25 − 1) + 4t^(log10 4 − 1) − 19) = 0.

Dank2 · Sep 6, 2019

Answers are x=0, x=-1.
Anyone have a clue how to show it ? it doesn't have to be a proof.

aheight · Sep 8, 2019

Dank2 said:

Answers are x=0, x=-1.
Anyone have a clue how to show it ? it doesn't have to be a proof.

If the equation is:
$$\log_{10}\big[25^{x+1}+4^{x+1}-19\cdot 10^x\big]=\log_{10}\big[10^{x+1}\big]$$,
then with some algebra, you can reduce it down to:
$$
25^{x+1}+4^{x+1}=\frac{29}{10}\cdot 10^{x+1}
$$
and can easily see ##x=0## is a solution and if you plug in ##x=-2##, confirm -2 is a solution also.

Exponential equation with logs

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