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

Question

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?

Answer 1

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

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

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

Answer 2

Rewrite

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

Then after simplification,

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

Answer 3

we have $$2^{2x}\cdot 5^{4x}\cdot 5^3=5^{2x}\cdot 10^3$$ from hwer we have $$5^{2x}=2^3$$