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I've got an equation I need to solve for $x$ but for the life of me can't work out how the answer is

$$x = \log_{25}8 $$

The original equation is

$$4^x \cdot 5^{4x+3} = 10^{2x+3}$$

Could someone please point me in the right direction?

David E
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    Checking with a calculator, your given value of $x$ does not seem to solve the equation. – paw88789 Mar 29 '17 at 12:48
  • I made a mistake where it should have been multiply not add the two exponential parts together. – David E Mar 29 '17 at 12:55
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    @mrnovice provided an answer to this corrected problem that he deleted because he misread what you first wrote as what you intended to write! I hope he reinstates it. – Ethan Bolker Mar 29 '17 at 12:55
  • @EthanBolker: yes, but the answer he provided was and is still wrong as it assumes a $+$. –  Mar 29 '17 at 13:01
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    @YvesDaoust Yes, his answer was literally wrong. But it was wrong precisely because the problem was wrong and he misread it as what it ought to have been. That was sensible and reasonable, and led to the actual error. So he should repost the answer and correct it and get double credit for reading the OP's mind. – Ethan Bolker Mar 29 '17 at 13:05
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    @EthanBolker: Using the identity $\log(a+b)=\log a+\log b$ in any context is a crime against logarithms, which should be prosecuted exponentially. –  Mar 29 '17 at 13:16
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    @YvesDaoust that made me lol – mrnovice Mar 29 '17 at 13:18

3 Answers3

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$4^x\cdot 5^{4x+3} = 10^{2x+3}$

$\log(4^x \cdot 5^{4x+3}) = \log(10^{2x+3})$

We now use the rules $\log(a^b) = b\log a$ and $\log(ab) = \log a + \log b$

$x\log 2^2 +(4x+3)\log 5=(2x+3)\log(5\cdot2)$

$2x\log 2+(4x+3)\log5 = (2x+3)(\log 5+\log 2)$

$(2x-2x)\log 2 + (4x-2x)\log 5 = 3(\log 5+\log 2)-3\log 5$

$2x\log 5 = 3\log 2$

$x = \frac{3\log 2}{2\log 5} $=$ \frac{\log 8}{\log25}$

Now note the change of base formula:

$\log_{b}x = \frac{\log_{a}x}{\log_{a}b}$

$\implies x = \log_{25}8$ as required

mrnovice
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    You seem to have also used the "rule" $\log(a+b) = \log(a)+\log(b)$ which is very very wrong. – 5xum Mar 29 '17 at 12:46
  • This makes sense, sorry about the confusion originally. I was going $log10 = 1$ rather than splitting it up. Thanks! – David E Mar 29 '17 at 13:05
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we have $$2^{2x}\cdot 5^{4x}\cdot 5^3=5^{2x}\cdot 10^3$$ from hwer we have $$5^{2x}=2^3$$

Dr. Sonnhard Graubner
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Rewrite

$$2^{2x}\cdot125\cdot25^{2x}=1000\cdot10^{2x}.$$

Then after simplification,

$$\left(\frac{2\cdot25}{10}\right)^{2x}=\frac{1000}{125},\\ 25^x=8.$$