For a>0 and a 1, au=av gives u=v: an exponential function is one-to-one. A base between 0 and 1 also gives a one-to-one function. Method: rewrite every number as a power of one positive base; use index laws; equate exponents; solve; substitute into the original equation. For 8x-1=4x+1, write 23x-3=22x+2. Thus 3x-3=2x+2, so x=5. For shifted powers added together, factor the common exponential term first: 3x+1+3x=4(3x). The rule au+av=au+v is false. Exam check: ax>0 for every real x, so an equation ax=0 or ax<0 has no real solution. If the equation is quadratic in ax, use the separate reducible-equations skill and keep ax>0.

For \(a>0\) and \(a\ne1\), \(a^u=a^v\) gives \(u=v\): an exponential function is one-to-one. A base between \(0\) and \(1\) also gives a one-to-one function.

Method: rewrite every number as a power of one positive base; use index laws; equate exponents; solve; substitute into the original equation.

For \(8^{x-1}=4^{x+1}\), write \(2^{3x-3}=2^{2x+2}\). Thus \(3x-3=2x+2\), so \(x=5\).

For shifted powers added together, factor the common exponential term first: \(3^{x+1}+3^x=4(3^x)\). The rule \(a^u+a^v=a^{u+v}\) is false.

Exam check: \(a^x>0\) for every real \(x\), so an equation \(a^x=0\) or \(a^x<0\) has no real solution. If the equation is quadratic in \(a^x\), use the separate reducible-equations skill and keep \(a^x>0\).

Matching exponential bases

Matching exponential bases