# Solving logarithmic problem

How to solve this:

log(base16)x + log(base4)x + log(base2)x = 7

If I have log(base16)x for example and i make

10^(log(base16)x)

of it, can I transform the base 16 into an exponent?

It would look like that then:

10^(logx)^16 = x^16

Would that be correct?

Then, I could solve it:

10^(logx)^16 * 10^(logx)^4 * 10^(logx)^2 = 10^7

that would be

x^16 * x^4 * x^2 = 10^7

x^22 = 10^7
22 = log(base x)10^7

Is that correct? Can I solve it this way?

I don't (think) your method is valid, but I think using Latex in your post would help clarify your idea. Here's the normal way you tackle the problem:

$$\log _{16}x+\log_{4}x+\log_{2}x=7$$

If you ever calculated non-natural logs on your calculator, you know that...

$$\log_{B}x = \frac{\ln x}{\ln B}$$

Replace all left terms by their respective fraction:

$$\frac{\ln x}{\ln 16}+\frac{\ln x}{\ln 4}+\frac{\ln x}{\ln 2}=7$$

Recall that ##\ln(a^b) = b\ln a##, and that ##16## and ##4## are powers of ##2##.

$$\frac{\ln x}{\ln 2^4}+\frac{\ln x}{\ln 2^2}+\frac{\ln x}{\ln 2}=7$$
$$\frac{\ln x}{4\ln 2}+\frac{\ln x}{2\ln 2}+\frac{\ln x}{\ln 2}=7$$

Find the common denominator, and consolidate the expression into one fraction:

$$\frac{\ln x}{4\ln 2}+\frac{2\ln x}{4\ln 2}+\frac{4\ln x}{4\ln 2}=7$$
$$\frac{7\ln x}{4\ln 2}=7$$

Divide by ##7## to cancel them out, and multiply each side by ##4\ln 2## to isolate \ln x.

$$\ln x=4\ln 2$$

Recall that ##4\ln 2 = \ln 2^4 = \ln 16##:

$$\ln x = \ln 16.$$

Raise both sides to e to cancel out the logarithms, and you find that:

$$x = 16$$

Last edited:
PFuser1232
Integral
Staff Emeritus