Answer :
Here is my written solution, I hope it helps you :)
Alternatively to this you could use the equation of 7500 x 1.06^t and use trial and error by replacing the value of t until you go beyond £10000.
Alternatively to this you could use the equation of 7500 x 1.06^t and use trial and error by replacing the value of t until you go beyond £10000.

Answer: 5 years
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Work Shown:
[tex]A = P*(1+r/n)^{n*t}\\\\10,000 = 7500*(1+0.06/1)^{1*t}\\\\10,000/7500 = (1.06)^{t}\\\\1.33333 \approx (1.06)^{t}\\\\\text{Log}(1.33333) \approx \text{Log}\left((1.06)^{t}\right)\\\\\text{Log}(1.33333) \approx t*\text{Log}(1.06)\\\\t \approx \frac{\text{Log}(1.33333)}{\text{Log}(1.06)}\\\\t \approx 4.937[/tex]
The first equation shown is the compound interest formula. I'm assuming that the bank is compounding annually (which means n = 1).
Whenever we have an exponent we want to solve for, we'll involve logs. A good phrase to remember is "If the exponent is in the trees, then log it down".
It takes approximately t = 4.937 years for the £7500 to become £10,000.
Round up to the nearest year to get t = 5 so we ensure that Brian has more than £10,000.