Question

please solve this
5^logx - 3^logx - 1 = 3^logx + 1 - 5^logx-1​

Answer 1

Answer:

The given equation is

5^{\log x}+3^{\log x}=3^{\log x+1}-5^{\log x-1}5logx+3logx=3logx+1−5logx−1

Let as assume log x = t. So, the given equation can be written as

5^{t}+3^{t}=3^{t+1}-5^{t-1}5t+3t=3t+1−5t−1

5^{t}+5^{t-1}=3^{t+1}-3^{t}5t+5t−1=3t+1−3t

Using product and quotient property of exponent.

5^{t}+\frac{5^{t}}{5}=3^{t}\cdot 3-3^{t}5t+55t=3t⋅3−3t [\because a^{m+n}=a^ma^n,\frac{a^m}{a^n}=a^{m-n}][∵am+n=aman,anam=am−n]

5\cdot 5^{t}+\frac{5^{t}}{5}=3^{t}\cdot 3-3^{t}5⋅5t+55t=3t⋅3−3t

5^{t}(1+\frac{1}{5})=3^{t}(3-1)5t(1+51)=3t(3−1)

5^{t}(\frac{6}{5})=3^{t}(2)5t(56)=3t(2)

Isolate the variable terms on one side.

\frac{5^t}{3^t}=\frac{5}{6}\times 23t5t=65×2

(\frac{5}{3})^t=\frac{5}{3}(35)t=35

It can be written as

(\frac{5}{3})^t=(\frac{5}{3})^1(35)t=(35)1

On comparing both sides we get

t=1t=1

log x= 1logx=1 [\because t=\log x][∵t=logx]

log x= log 10logx=log10 [\because \log 10=1][∵log10=1]

x=10x=10

Therefore, the value of x is 10.