So we have the equation initially in the form 3^(5x-2)=4^(6-x), and as the solution involves log10, then a sensible first move would be to take log10 of both sides, giving log10(3^(5x-2)) = log10(4^(6-x)).

Using the log law that loga(k) = kloga, we can rewrite this as (5x-2)log10(3) = (6-x)log10(4).

Then expanding the brackets, and taking the x values to one side and the constants to the other gives, 5xlog10(3) + xlog10(4) = 6log10(4) + 2log10(3).

Taking x out as a factor, we get x(5log10(3) + log10(4)) = 6log10(4) + 2log10(3).

Using the log law we made use of earlier in the question (loga(k) = kloga), this can be written as x(log10(243) + log10(4)) = log10(4096) + log10(9).

We can then use the log law, loga(b) + loga(c) = loga(bc), to write the equations as x(log10(972)) = log10(36864).

Now all we have to do is divide by log10(972), and we have our answer, so,

x = log10(36864) / log10(972).

differentiate 4x^3 + 3x

Related Maths A Level answers

Solve the equation 3^(5x-2)=4^(6-x), and show that the solution can be written in the form log10(a)/log10(b).

Integrate 2x/(x^2+3) using the substitution u=x^2+3

Use simultaneous equations to find the points where the following lines cross: 3x - y = 4 and x^2 + 7y = 5

How do I solve a quadratic equation?