Expert: Josh - 1/20/2007

Question
Hi, I am looking for some help in solving for "x" in the following 2 equations:

265(1+x)^2 = 307

625(1.08)^x = 1284

Thanks!

Answer
No problem Jake.

The key thing to remember is that an exponential function is inverted by a logarithmic function.

e.g., Consider the expression z-squared. If z^2 = y, then, z=log_2(z^2)=log_2(y).

For 265(1+x)^2 = 307, we first rewrite this as
(1+x)^2=307/265. To undo the exponent, we take the logarithm (of base two) on both sides of the equation.

This gives (1+x)=log_2(307/265), from this, we easily obtain x = log_2(307/265)-1.

For the second question, we have 1.08^x=(1284/625).
We use the following trick:
First, recognize that a^x=exp(log_e(a^x)), where log_e (sometimes written log or ln) is the natural logarithm defined with respect to base "e"~=2.718281...

Applying the logarithmic rule, log(a^x)=x*log(a), you can verify this yourself, we have a^x=exp(x*ln(a)). [Note: Plug in some numbers on your calc if you are not convinced]

Let a^x=b. Your question is in the same form, with a=1.08, b=1284/625. We can now make x the subject, by taking the natural logarithm "ln" on both sides of the equation, exp(x*ln(a))=b.

We get x*(ln(a))=ln(b). Hence x=ln(b)/ln(a).

Numerically, a=1.08, b=1284/625, x=log(b)/log(a)=9.35517274564933.

Let me know if anything is unclear.