Question

What is the exact solution to the equation 2^2x=5^x-1? In Log Form.

Answer 1

2^2x=5^x-1 (x^n=x^m=n=m)

2x=5x-1

2x-5x=-1

-3x=-1

3x=1

x=1/3

Answer 2

$x = \dfrac{\log 5}{\log \frac{5}{4} }$

Step-by-step explanation:

We have to get the exact solution to the equation $2^{2x} = 5^{(x - 1)}$ in log form.

Now, $2^{2x} = 5^{(x - 1)}$

Taking log on both sides we get,

$\log 2^{2x} = \log 5^{(x - 1)}$

⇒ 2x log 2 = (x - 1) log 5

{Since, we know the property of logarithm that $\log a^{b} = b \log a$ }

⇒ x log 5 - 2x log 2 = log 5

⇒ x(log 5 - 2 log 2) = log 5

⇒ $x(\log 5 - \log 2^{2}) = \log 5$

{Since, we know the property of logarithm that $\log a^{b} = b \log a$ }

⇒ $x(\log \frac{5}{2^{2}}) = \log 5$

{Since, we have the logarithmic property (log a - log b) = log (a/b)}

⇒ $x\log \frac{5}{4} = \log 5$

⇒ $x = \dfrac{\log 5}{\log \frac{5}{4} }$ (Answer)