log-4-5-x-3-x-log-5-4-x-3-x-

Question Number 198222 by cortano12 last updated on 14/Oct/23

\log_{4} (5^{x} - 3^{x}) = \log_{5} (4^{x} + 3^{x})

Commented by Rasheed.Sindhi last updated on 14/Oct/23

4^{2} + 3^{2} = 5^{2}

\log_{5} (4^{2} + 3^{2}) = \log_{5} 5^{2}

\log_{5} (4^{2} + 3^{2}) = 2 \dots \dots (i)

5^{2} - 3^{2} = 4^{2}

\log_{4} (5^{2} - 3^{2}) = \log_{4} 4^{2}

\log_{4} (5^{2} - 3^{2}) = 2 \dots \dots (i i)

(i) & (i i) :

\log_{5} (4^{2} + 3^{2}) = \log_{4} (5^{2} - 3^{2})

Comparing with

\log_{5} (4^{x} + 3^{x}) = \log_{4} (5^{x} - 3^{x})

yields

x = 2

Answered by Rasheed.Sindhi last updated on 14/Oct/23

\log_{4} (5^{x} - 3^{x}) = \log_{5} (4^{x} + 3^{x}) = k

4^{k} = 5^{x} - 3^{x}, 5^{k} = 4^{x} + 3^{x}

4^{k} + 5^{k} = 5^{x} + 4^{x} \Rightarrow k = x

∴ \log_{4} (5^{x} - 3^{x}) = \log_{5} (4^{x} + 3^{x}) = x

4^{x} = 5^{x} - 3^{x}, 5^{x} = 4^{x} + 3^{x}

Both are equivalent to :

4^{x} + 3^{x} = 5^{x}

Which is only true for

x = 2