Question

1. solve exponential equations

if ( a ^ { u } = a ^ { v } ), then ( u = v ).
for example, if ( 3 ^ { x } = 3 ^ { 7 } ) then
this means that if you can rewrite your terms to have the same base, then your exponents are equal to each other.
example 1:
solve ( 3 ^ { x + 1 } = 81 )
( 81 = 3 ^ { ? } )
( 3 ^ { x + 1 } = 81 = 3 ^ { ? } )
if ( 3 ^ { x + 1 } = 3 ^ { ? } ) then ( x = )
example 2:
solve ( 4 ^ { 2 x - 1 } = 8 ^ { x + 3 } )
( 4 ^ { 2 x - 1 } = )( = )( = 2 ^ { ? } )
( 8 ^ { x + 3 } = )( = )( = 2 ^ { ? } )

Explanation:

Step1: Solve \(3^x = 3^7\)

According to the rule \(a^u=a^v\Rightarrow u = v\), when \(a = 3\), \(u=x\), \(v = 7\), we have \(x=7\).

Step2: Solve \(3^{x + 1}=81\)

Since \(81=3^4\), the equation \(3^{x + 1}=81\) becomes \(3^{x+1}=3^4\).
Using the rule \(a^u=a^v\Rightarrow u = v\) (\(a = 3\), \(u=x + 1\), \(v = 4\)), we get \(x+1=4\).
Subtract \(1\) from both sides: \(x=4 - 1=3\).

Step3: Solve \(4^{2x-1}=8^{x + 3}\)

Rewrite \(4\) as \(2^2\) and \(8\) as \(2^3\).
For \(4^{2x-1}=(2^2)^{2x-1}\), using the power - of - a - power rule \((a^m)^n=a^{mn}\), we have \((2^2)^{2x-1}=2^{2(2x - 1)}=2^{4x-2}\).
For \(8^{x + 3}=(2^3)^{x + 3}\), using the power - of - a - power rule \((a^m)^n=a^{mn}\), we have \((2^3)^{x + 3}=2^{3(x + 3)}=2^{3x+9}\).
Since \(4^{2x-1}=8^{x + 3}\), then \(2^{4x-2}=2^{3x + 9}\).
Using the rule \(a^u=a^v\Rightarrow u = v\) (\(a = 2\), \(u=4x-2\), \(v = 3x + 9\)), we set \(4x-2=3x + 9\).
Subtract \(3x\) from both sides: \(4x-3x-2=3x-3x + 9\), which gives \(x-2=9\).
Add \(2\) to both sides: \(x=9 + 2=11\).

Answer:

For \(3^x = 3^7\), \(x = 7\).
For \(3^{x + 1}=81\), \(x = 3\).
For \(4^{2x-1}=8^{x + 3}\), \(x = 11\).